( Lateral Loads ) Seismic &Wind
By: Eng. Ahmed Farghal.
Table of Contents.
Introduction.
Equivalent static load method.
Steps of Calculating Seismic Load.
Wind Loads.
Systems resisting lateral loads.
Design of Shear wall.
Drift of structures due to seismic loads
Examples.
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Page No.
2
12
25
37
48
57
66
70
Lateral Loads
Introduction.
1- Seismic loads.
2- Wind loads.
1. Seismic Loads:
(
)
Seismic sources
1- Movements of tectonic plates
2- Movements of faults
3- Volcanoes
4- Failure of roof of large cave
5- Mankind effect (explosion, fill and in-fill of dams ..... etc.)
6- Undefined reasons
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Page No.
Types of surface waves
Love wave
Rayleigh wave
Love wave
Di
re c
Pa
rti
cle
Rayleigh wave
tio
Di
no
re c
fp
rop
ag
ati
on
Pa
rti
mo
tio
n
cle
tio
no
mo
tio
n
Classification of earthquakes
1- Deep focus earthquakes
Focal depth > 300 km
Epicentral
distance
Epicenter
2- Intermediate focus earthquakes
300 km > Focal depth > 70 km
Focal
depth
3- Shallow focus earthquakes
Focal depth < 70 km
l
tra
n
ce nce
o
yp ta
H dis
Focus
or Hypocenter
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Page No.
fp
rop
ag
ati
on
Methods of measuring earthquakes magnitude
The Richter Scale
- The magnitude of most earthquakes is measured on richter scale.
- It was invented by Charles F. Richter in 1934.
- The richter magnitude is calculated from the amplitude of the
largest seismic wave recorded for the earthquake, no matter what
type of wave was the strongest.
The Mercalli Scale
- It is another way to measure the strength of an earthquake.
- It was invented by Giuseppe Mercalli in 1902.
- This scale uses the observations of the people who experienced
the earthquake to estimate its intensity.
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Page No.
Methods of analysis of structures under seismic load
1- Equivalent Static Load (Simplified Modal Response Spectrum).
2- Multi-Modal Response Spectrum Method.
3- Time History Analysis.
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Page No.
1- Equivalent Static Load (Simplified Modal Response Spectrum).
Fundamental period (T1)
T1 < 4 T C
where:
TC =
1 H
and
T1 < 2.0 Seconds
Response Spectrum Curve
60 m
H =
H
2
L
B
4.0
B
L =
B=
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H
)
Page No.
L
3 Uniform shape
4 Uniform statical system
Sec 4
Sec 3
Sec 2
H
Sec 1
Core
Sec 2
Shear wall
Sec 3
Solid Slab
Sec 4
Flat Slab
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Sec 1
Page No.
(Lx / L y )
(e o )
(x , y
e > 0.15 L
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Page No.
L2
L1
H
L3
L1
L
L1 - L 2
< 0.20
L1
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> 0.15 H
L1 + L 3
< 0.20
L
)
Page No.
L2
L1
H
L3
L1
L
L
< 0.15 H
L 1+ L3
< 0.50
L
L - L2
< 0.30
L
L1 - L 2
< 0.10
L1
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2- Multi-Modal Response Spectrum Method
Sd (T)
Response Spectrum
Curve
Time (Sec)
( Rotation
Response Spectrum Curve
Displacement
3- Time History Analysis.
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Page No.
Equivalent static load method:
(Simplified Modal Response Spectrum Method)
Z
Fb
Fb
1
L
B
Side View
Y
1
L
Plan
Fb
X
re
Mo tical
Cri
B
Elevation
Fb
2
( More Critical )
2
X ,Y
Manual
More Critical
Y
X
0.3 E Fx
E Fx
E Fy
0.3 E Fy
ET = 0.3 E Fx + E Fy
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ET = E Fx + 0.3 E Fy
)
Page No.
+
Fb = Sd (T1)
l W
g
where:
Fb
= Ultimate base shear force
= Gravitational acceleration
g
Sd (T1) = Response Spectrum
(T1)
Sd (T)
2.5 ag S g1
Response Spectrum
Curve
ag S g1
Sd (T)
1
TB TC
T1
( Rotation
TD
4.0
Time (Sec)
Response Spectrum Curve
Displacement
Response Spectrum
0 < T < TB
2.5 h - 2 )
: Sd (T) = a g g1 S 32 + T
(
TB
R
3
TB < T < TC
h
: Sd (T) = a g g1 S 2.5
R
TC < T < TD
: Sd (T) = a g g1 S 2.5
R
TC h
> 0.20 a g g1
T
TC TD h
TD < T < 4.0 Sec : Sd (T) = a g g1 S 2.5
>
R
T2
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0.20 ag g1
Page No.
where:
T
= Periodic time of different mode shapes
(Different mode shapes)
TB,TC ,TD , S , a g
ag =
Zone
Design acceleration
Zone 1
Zone 2
Zone 3
Zone 4
Zone 5A
Zone 5B
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(a g )
0.10 g
0.125 g
0.15 g
0.20 g
0.25 g
0.30 g
)
Page No.
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26
28
T
ER
S
E
TD
27
29
Dakhla
Oasis
Bahariya
Oasis
30
31
Kharga
Oasis
Asyut
El Minya
32
Sohag
Beni Suef
Cairo
Giza
Ismailia
Damietta
32
33
Kosetta
Port Said
Damanhur
El Mansura
Tanta
Al Fayum
Bahariya
Alamein
Alexandria
31
33
Aswan
35
Ras Muhammad
Quseir
Safaga
Hurghada
34
Kom Ombo
Idfu
Luxor
Qena
El Arish
34
36
Page No.
35
36
22
37
23
24
25
26
27
28
29
30
31
37
32
Halaib
Ras Banas
Marsa Alam
RT
E
ES
D
T
22
25
S
WE
23
30
S
EA
24
25
Siwe
Matruh
29
MEDITERRIAN SEA
28
EZ
SU
26
27
28
29
30
31
Salum
27
G
26
2
) ces/ mc( AGP
25
32
L
FLU
GU
F
FO
AF O
AI
AB A
U
S IN
EA
S
D
RE
250
200
150
125
100
50
Zone 5B
Zone 5A
Zone 4
Zone 3
Zone 2
Zone 1
T B ,TC ,TD , S
TB ,TC = elastic response spectrum
TD
= spectrum
S
= Soil Factor
Subsoil Soil
Class Type
A
Rock
B
Dense
Soil
Soil Description
Meduim
Soil
C
D
Loose
Soil
For Type (1)
Subsoil Class
A
B
C
D
S
1.00
1.35
1.50
1.80
TB
0.05
0.05
0.10
0.10
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TC
0.25
0.25
0.25
0.30
)
TD
1.20
1.20
1.20
1.20
Page No.
For Type (2)
Subsoil Class
A
B
C
D
g1 =
S
1.00
1.20
1.25
1.35
TB
0.15
0.15
0.20
0.20
TC
0.40
0.50
0.60
0.80
TD
2.00
2.00
2.00
2.00
Importance factor
Importance
Category
Type of Structures
Importance
Factor (g1)
I
1.40
II
1.20
III
1.00
IV
0.80
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Page No.
R = Response modification factor
R
:
4.50
3.50
2.00
:
5.00
4.50
4.50
:
7.00
5.00
6.00
5.00
:
:
3.00
3.50
5.00
Response modification factor (R)
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3.00
3.50
6.00
5.00
7.00
5.00
5.00
4.50
4.50
4.50
3.50
2.00
R
5.00
:
:
:
Response modification factor (R)
:
:
R = Response modification factor
+
R.C. Shear Walls or Cores
Frames with Bracing
OR
R = 5.0
R = 4.5
Non Ductile Frames
R = 5.0
Ductile Frames
R = 6.0
Non Ductile Frames
R = 5.0
Ductile Frames
R = 7.0
NO R.C. Shear Walls
or Cores or Bracing
R.C. Shear Walls or Cores
Frames with Bracing
R
h=
Damping factor corrected for horizontal response spectrum
Type of Structure
Steel with Welded Connections
Steel with Bolted Connections
Reinforced Concrete
Prestressed Concrete
Reinforced Masonry Walls
1.00
h
1.20
1.05
1.00
1.05
0.95
Damping factor (h)
Prestressed Concrete
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Page No.
Response Spectrum
+
Fb = Sd (T1)
Fb
l W
g
where:
Fb
= Ultimate base shear force
= Gravitational acceleration
g
Sd (T1) = Response Spectrum
(T1)
T1 = Fundamental period of the building
T1 = Ct H
3/4
where:
H = Height of the building from foundation level
H
Ct = Factor depend on structural system and material
Structural System
Steel moment resisting frames
Reinforced concrete moment resisting frames
(Space frames)
Ductile frames (beams & columns)
Non-ductile frames (flat slabs)
All other buildings
Cores or Shear walls
Combinations of (cores or shear walls)
& frames
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Ct
0.085
0.075
0.050
Page No.
Shear wall + Core ( Ct = 0.05)
Frames + Shear wall (C t = 0.05)
Non-ductile frames (Flat Slab)
( C t = 0.075)
Ductile frames (Beams + Slab)
( C t = 0.075)
T1
Range
Response Spectrum
0 < T < TB
Sd (T1)
T ( 2.5 h - 2 )
: Sd (T) = a g g1 S 32 + T
R
3
B
TB < T < TC
h
: Sd (T) = a g g1 S 2.5
R
TC < T < TD
: Sd (T) = a g g1 S 2.5
R
TC h
> 0.20 a g g1
T
TC TD h
TD < T < 4.0 Sec : Sd (T) = a g g1 S 2.5
> 0.20 ag g1
2
R
T
TB ,TC ,TD , S , a g
5.00
Response modification factor (R)
1.00
Damping factor (h)
Prestressed Concrete
Importance factor (g1)
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Page No.
l = Correction factor
If T1 < 2 TC
l = 0.85
If T1 > 2 TC
l = 1.00
W = Total weight of the structure above foundation level
W = S (w i )
wi = ( i )
+
+
+
+
wi = ( D.L. + a L.L.) Floor Area + O.W. of beams and columns
wi = ( g s + a Ps ) Floor Area + O.W. of beams and columns
a=
a
0.25
0.50
1.00
NOTE
- D.L. ( g s ) & L.L. ( Ps ) are working loads
( working loads ) ‫ﻮن‬
g s = t s gc + F.C. + walls
2
If given in kN/m
P s = L.L.
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‫ﺟﻤﯿ ﻊ اﻷﺣﻤ ﺎل ﯾﺠ ﺐ أن ﺗﻜ‬
Page No.
NOTE
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Page No.
Steps of Calculating Seismic Load:
+
Fb = Sd (T1)
1 Calculate
T1 = Ct H
3/4
l W
g
Structural System
Reinforced concrete moment resisting frames
(Space frames)
Ductile frames (beams & columns)
Non-ductile frames (flat slabs)
H
All other buildings
Cores or Shear walls
Combinations of (cores or shear walls)
& frames
Ct
0.075
0.050
2
Subsoil Soil
Class Type
3 Sd (T1 )
A
Rock
B
Dense
Soil
C
Meduim
Soil
D
Loose
Soil
R , h ,g1 ,TB ,TC ,TD , S , a g
For Type (1)
Subsoil Class
A
B
C
D
Importance
Category
Soil Description
For Type (2)
S
1.00
1.35
1.50
1.80
TB
0.05
0.05
0.10
0.10
TC
0.25
0.25
0.25
0.30
Type of Structures
TD
1.20
1.20
1.20
1.20
Importance
Factor (g1 )
I
1.40
II
1.20
III
1.00
IV
0.80
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Subsoil Class
A
B
C
D
Zone
Zone 1
Zone 2
Zone 3
Zone 4
Zone 5A
Zone 5B
S
1.00
1.20
1.25
1.35
TC
0.40
0.50
0.60
0.80
Design acceleration
(a g )
0.10 g
0.125 g
0.15 g
0.20 g
0.25 g
0.30 g
1.00
5.00
)
TB
0.15
0.15
0.20
0.20
(h)
(R)
Page No.
TD
2.00
2.00
2.00
2.00
T1
4
Range
Sd (T1)
T ( 2.5 h - 2 )
: Sd (T) = a g g1 S 23 + T
R
3
B
0 < T < TB
TB < T < TC
h
: Sd (T) = a g g1 S 2.5
R
TC < T < TD
: Sd (T) = a g g1 S 2.5
R
TC h
>
0.20 a g g1
T
TC TD h
TD < T < 4.0 Sec : Sd (T) = a g g1 S 2.5
>
R
T2
5 Get l = 0.85
T1 < 2 TC
l = 1.00
T1 > 2 TC
6 Calculate
0.20 ag g1
W
W = S (w i )
wi = ( i )
+
+
+
+
wi = ( D.L. + a L.L.) Floor Area + O.W. of beams and columns
wi = ( g s + a Ps ) Floor Area + O.W. of beams and columns
a
0.25
0.50
1.00
NOTE
- D.L. ( g s ) & L.L. ( Ps ) are working loads
( working loads ) ‫ﻊ اﻷﺣﻤ ﺎل ﯾﺠ ﺐ أن ﺗﻜ ﻮن‬
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‫ﺟﻤﯿ‬
Page No.
Distribution of lateral force on each floor
( In Elevation)
( In Plan)
sw 1
sw 2
F1-5
sw 3
Fb
F1-4
sw 4
F5
F4
F3
F2-5
F2-4
F1-3
F2-3
F1-2
F2-2
F1-1
F2
F2-1
sw1
F1
sw2
F3-5
F4-5
F3-4
+
Fi = Fb
w i Hi
wi Hi
F4-4
F3-3
F4-3
F3-2
i=n
F4-2
F3-1
i=1
F4-1
sw3
where:
Fi = ( i )
sw4
h8
Fb = Ultimate base shear force
+
Fb = Sd (T1)
Hi =
h7
h6
l W
g
H5
H4
‫ ( ﻣﻘﺎﺳ ﺎً ﻣﻦ ﻣﻨﺴ ﻮب اﻷﺳﺎﺳ ﺎت‬i ) ‫ارﺗﻔ ﺎع ﺑﻼﻃ ﺔ اﻟ ﺪور رﻗ ﻢ‬
H3
H2
H1
wi = ( i )
h5
h4
h3
h2
h1
+
+
+
+
wi = ( D.L. + a L.L.) Floor Area + O.W. of beams and columns
wi = ( g s + a Ps ) Floor Area + O.W. of beams and columns
D.L. ( g s ) & L.L. ( Ps ) are working loads
( working loads ) ‫ﻊ اﻷﺣﻤﺎل ﯾﺠ ﺐ أن ﺗﻜ ﻮن‬
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‫ﺟﻤﯿ‬
Page No.
F8
F7
F6
F5
F4
F3
Fb
F2
F1
Total seismic load
Story Forces
F8
h8
F7
h7
F6
h6
F5
h5
H5
F4
h4
H4
F3
h3
H3
F2
h2
H2
F1
H1
h1
Load Diagram
Q 8 = F8
Q 7 = Q 8+ F7
+
M8 = Q 8
Q 6 = Q 7+ F6
+
M7 = M 8 + Q 7 h 7
Q 5 = Q 6+ F5
+
M6 = M 7 + Q 6 h 6
Q 4 = Q 5+ F4
+
M5 = M 6 + Q 5 h 5
Q 3 = Q 4+ F3
+
M4 = M 5 + Q 4 h 4
Q 2 = Q 3+ F2
+
M3 = M 4 + Q 3 h 3
Q 1 = Q 2+ F1
M2 = M 3 + Q 2 h 2
+
Fb = S F i
h8
+
Mbase U.L.= M 2 + Q1 h 1
Overturning Moment
Shear Diagram
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+
+
+
+
+
Mbase U.L.= M2 + Q 1 h 1
=S Fi H i = F1 H 1 + F2 H 2 + F3 H 3 + ...............
)
Page No.
wi (kN)
Hi (m)
wi Hi
+
Fi (kN)
w8
H8
w8 H 8
+
F8
Q8 = F8
M8 = Q 8 h 8
7
w7
H7
w7 H 7
+
F7
Q 7 = Q 8 + F7
M7 = M 8 + Q 7 h 7
6
w6
H6
w6 H 6
+
F6
Q 6 = Q 7 + F6
M6 = M 7 + Q 6 h 6
5
w5
H5
w5 H 5
+
F5
Q 5 = Q 6 + F5
M5 = M 6 + Q 5 h 5
4
w4
H4
w4 H 4
+
F4
Q 4 = Q 5 + F4
M4 = M 5 + Q 4 h 4
3
w3
H3
w3 H 3
+
F3
Q 3 = Q 4 + F3
M3 = M 4 + Q 3 h 3
2
w2
H2
w2 H 2
+
F2
Q 2 = Q 3 + F2
M2 = M 3 + Q 2 h 2
1
w1
H1
w1 H 1
F1
Q 1 = Q 2 + F1
M base U.L.= M 2 + Q 1 h 1
S wi H i
S Fi
SQ
where:
Floor No. = ‫ﺬ‬
i
+
SH
M i (kN.m)
+
+
+
+
+
+
+
Q i (kN)
+
+
Floor
No.
8
i
‫ﺎت و ﻟﯿ ﺲ ﻟ ﻸدوار ﻟ ﺬﻟﻚ ﻓ ﻲ ﺣﺎﻟ ﺔ وﺟﻮد ﺑ ﺪروم ﺗﺄﺧ‬
‫ﺮﻗﯿﻢ ﻟﻠﺒﻼﻃ‬
‫ھﻮ ﺗ‬
( ) ‫ﺑﻼﻃ ﺔ اﻟﺒ ﺪروم رﻗ ﻢ‬
wi = ( i )
+
+
+
+
wi = ( D.L. + a L.L.) Floor Area + O.W. of beams and columns
wi = ( g s + a Ps ) Floor Area + O.W. of beams and columns
h8
h7
Hi =
wi H i =
+
h6
‫ ( ﻣﻘﺎﺳ ﺎً ﻣﻦ ﻣﻨﺴ ﻮب اﻷﺳﺎﺳ ﺎت‬i ) ‫ارﺗﻔ ﺎع ﺑﻼﻃ ﺔ اﻟ ﺪور رﻗ ﻢ‬
H5
w i Hi
w i Hi
‫ﻹﯾﺠ ﺎد اﻟﻤﻌﺎﻣ ﻞ‬
i=n
h4
H4
‫و ﯾﺘ ﻢ ﺣﺴ ﺎﺑﮭﺎ‬w i & H i ‫ھﻮ ﺣﺎﺻ ﻞ ﺿ ﺮب اﻟﻌﻤ ﻮدﯾﻦ‬
h3
H3
h2
H2
H1
h1
i=1
w i Hi
=
w i Hi
i=n
‫ﺔ‬
‫ ﻋﻠ ﻰ اﻷدوار اﻟﻤﺨﺘﻠﻔ‬base shear (F ) ‫ﺒﺔ اﻟ ﺘﻰ ﯾﺘ ﻢ ﺑﮭ ﺎ ﺗﻮزﯾ ﻊ‬
b
‫و ھﻲ اﻟﻨﺴ‬
i=1
i=1
+
wi & wi H i - ‫ﻮدﯾﻦ‬
w i Hi
H
= i=n i
i=n
w i Hi
Hi
‫و ذﻟ ﻚ ﻷن‬
‫ﻓ ﻲ ﺣﺎﻟ ﺔ ﺗﺴ ﺎوي وزن اﻷدوار ﻻ داﻋﻲ ﻹﺿ ﺎﻓﺔ اﻟﻌﻤ‬
i=1
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h5
Page No.
Fi = ( i )
+
Fi = Fb
wi H i
i=n
wi Hi
i=1
+
Fb = Ultimate base shear force
Fb = Sd (T1) l W
g
Q i = ‫ﺰﻟﺰال‬
‫ﺔ ﻟﻠ‬
‫ﺄﺛﯿﺮ اﻟﻘ ﻮى اﻷﻓﻘﯿ‬
‫ﺔ ﻟﻠ ﺰﻟﺰال‬
Mi =
‫ﻗ ﻮى اﻟﻘ ﺺ اﻟﻨﺎﺗﺠ ﺔ ﻣﻦ ﺗ‬
‫ﺄﺛﯿﺮ اﻟﻘ ﻮى اﻷﻓﻘﯿ‬
‫اﻟﻌ ﺰوم اﻟﻨﺎﺗﺠ ﺔ ﻣﻦ ﺗ‬
‫ ﻋﻦ ﻃ ﺮﯾﻖ اﺳ ﺘﺨﺪام اﻟﻘ ﺎﻧﻮن اﻵﺗ ﻲ‬Q ‫ﺎب‬
i ‫ ﺑ ﺪون ﺣﺴ‬M
- ‫ﺎب‬
i ‫ﯾﻤﻜ ﻦ ﺣﺴ‬
+
+
+
+
Mbase U.L.=S Fi H i = F1 H 1 + F2 H 2 + F3 H 3 + ....................
Check Overturning
F8
+
Mbase U.L. S Fi H i
Moverturning = 1.40
= 1.40
+
Mstability = W
F7
F6
F5
F4
F3
B
2
Fb
F2
W
F1
where:
W = Total weight of structure
‫ﺄ‬
Moverturning
B
‫اﻟ ﻮزن اﻟﻜﻠ ﻲ ﻟﻠﻤﻨﺸ‬
Factor Of Safety = Stability Moment
Overturning Moment
F.O.S.
=
Mstability
< 1.5
Moverturning
‫( ﻛ ﺎﻵﺗﻲ‬M overturning ) - ‫ﺔ ﺣﺴ ﺎب‬
+
Mbase U.L. = ( Fb )
2H
3
‫و ذﻟ ﻚ ﺑﺸ ﺮط أن ﯾﻜ ﻮن وزن ﺟﻤﯿ ﻊ‬
Moverturning =
2H
3
‫ﺎً ﻋﻨ ﺪ‬
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‫ﺔ ﺗﺆﺛ ﺮ ﺗﻘﺮﯾﺒ‬
‫ﺔ ﺗﻘﺮﯾﺒﯿ‬
‫ﯾﻤﻜ ﻦ ﺑﻄﺮﯾﻘ‬
Mbase U.L.
1.40
‫ﺎر أن اﻟﻘ ﻮى اﻷﻓﻘﯿ‬
‫و ذﻟ ﻚ ﺑﺎﻋﺘﺒ‬
‫ﻣﺘﺴ ﺎوي‬. (wi) ‫اﻷدوار‬
)
Page No.
Check Sliding
Fb
1.40
Resisting Force = m W
where:
W = Total weight of structure
Sliding Force =
+
Fb
‫ﺄ‬
‫اﻟ ﻮزن اﻟﻜﻠ ﻲ ﻟﻠﻤﻨﺸ‬
m W
+
m
W
= Coefficient of friction
Factor Of Safety =
Resisting Force
< 1.5
Sliding Force
NOTE
Ultimate loads
Working loads
1.40
Mbase U.L. & Fb
Check sliding and Check overturning
Moverturning =
Fsliding
Mbase U.L.
1.40
Fb
=
1.40
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Page No.
Example.
A ten floor hospital located in Cairo with dimensions (20 x 40 m).
Height of each floor is 4.0 m. Soil below the building is very dense
sand and its coeff. of friction is 0.3. All floors are flat slab of average
thickness equal 0.3 m. Due to Earthquake loads , it is required to :
1- Calculate the ultimate base shear force .
2- Calculate the story shear and overturning moment at each floor
level and draw its distribution on the height of the building.
3- Find the bending moment and shearing forces acting at base
level of the building and draw distribution of shear forces.
4- Check The Stability of the building against silding and
overturning.
Given that :
4 +10 = 40 m
F.C. = 2.0 kN/m 2
Walls = 3.0 kN/m 2
L.L. = 3.0 kN/m 2
Elevation
40 m
20 m
0.30
Plan
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Page No.
Solution
1- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
Cairo
Map
Page (8)
Zone (3)
Very dense sand
Table
Page (9)
Table
Page (7)
a g = 0.15 g
Soil type (B)
Soil type (B)
Table
Page (9)
Response spectrum curve Type (1)
T1 = Ct H
S =
TB =
TC =
TD =
3/4
Shear wall
Table
Page (13)
C t = 0.05
Total Height of building (H) = 40 m
+
T1 = 0.05
40
3/4
= 0.795 sec.
Check
T1 < ( 4 TC = 1.0 sec.)
O.K.
Sd (T)
T1 < 2.0 sec.
O.K.
2.5 ag S g1
Response Spectrum
Curve
ag S g1
Sd (T)
1
TB TC
T1
TD
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4.0
)
Time (Sec)
Page No.
1.35
0.05
0.25
1.20
TC < T1 < TD
Page (14)
TC h
> 0.20 a g g1
T1
Sd (T1) = a g g1 S 2.5
R
R = 5.00
h = 1.00
Table
Page (10)
2.5
5.00
0.25
0.795
+
1.35
+
+
Sd (T1) = 0.15 g 1.40
+
g1 = 1.40
1.00
= 0.0446 g
0.20 a g g1 = 0.20
+
+
0.15 g 1.40 = 0.042 g < Sd (T1)
O.K.
T1 > 2 TC = 0.50 sec
l = 1.00
+
ws = D.L. + a L.L.
+
ws = ( t s gc + F.C. + Walls ) + a L.L.
Table
Page (15)
a = 0.50
+
+
+
+
+
ws = ( 0.30 25 + 2.0 + 3.0 ) + 0.50 3.0 = 14.0 kN/m2
wFloor = 14.0 20 40 = 11200 kN
wTotal = 11200 10 = 112000 kN
NOTE
t av
t av
l W
g
+
= 0.0446 g
1.0 112000
g
+
+
Fb = Sd (T1 )
Fb = 4995.2 kN
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Page No.
2- Distribution of lateral force on each floor
+
Fi = Fb
w i Hi
w i Hi
i=1
i=n
w i Hi
H
= i=n i
w i Hi i = 1 Hi
i=1
i=n
‫ﻷن وزن اﻟ ﺪور ﺛﺎﺑ ﺖ‬
i = 10
H = 40 + 36 + 32+ 28 +24+ 20 +16 + 12 + 8 + 4
i
i=1
= 220 m
Fi = ( 4995.2 ) 1 H i
220
36
28
20
Fi
=
16
22.705 H i
12
4
Floor H (m) F (kN) Q (kN)
i
i
i
No.
40.0 908.22 908.22
10
8
Mi (kN.m)
3632.87
9
36.0
817.40 1725.62
10535.33
8
32.0
726.57 2452.19
20344.09
7
28.0
635.75 3087.94
32695.85
6
24.0
544.93 3632.87
47227.35
5
20.0
454.11 4086.98
63575.27
4
16.0
363.29 4450.27
81376.35
3
12.0
272.47 4722.74 100267.29
2
8.0
181.64 4904.38 119884.80
1
4.0
90.82
4995.20 139865.60
Shearing force at base = 4995.2 kN
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Page No.
24
32
40
908.22
908.22
1725.62
817.40
3632.87
2452.19
726.57
10535.33
3087.94
635.75
20344.09
3632.87
544.93
32695.85
4086.98
454.11
47227.35
4450.27
363.29
63575.27
4722.74
272.47
81376.35
4904.38
181.64
100267.29
4995.20
90.82
119884.80
139865.60
Shear Diagram
Load Diagram
Moment Diagram
3- Check sliding
Fb
= 4995.2 = 3568 kN
1.40
1.40
Factor Of Safety =
W = 0.30 112000 = 33600
+
Resisting Force =m
+
Sliding Force =
Resisting Force
Sliding Force =
33600
3568
= 9.4 > 1.5
Safe
4- Check overturning
Mbase U.L. = 139865.6 kN.m
1.40
= 139865.6
= 99904 kN.m
1.40
+
Resisting Moment = WTotal
B
112000
2 =
+
Moverturning =
Mbase U.L.
Resisting Moment
Factor Of Safety = Over Turning Moment =
20 1120000 kN.m
2 =
1120000
11.2 > 1.5
99904 =
Safe
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Page No.
2. Wind Loads.
‫ھﻲ اﻟﻘ ﻮى اﻟ ﺘﻲ ﺗﺆﺛ ﺮ ﺑﮭ ﺎ اﻟﺮﯾ ﺎح ﻓ ﻲ اﺗﺠ ﺎه ﻣﺘﻌﺎﻣ ﺪ ﻋﻠ ﻰ أﺳ ﻄﺢ اﻟﻤﻨﺸ ﺄت‬
Ce = - 0.8
Pressure
(0.00)
Elevation
Wind load acting on the structure
kN/m
‫ﺿ ﻐﻂ اﻟﺮﯾ ﺎح اﻟﻤﺆﺛ ﺮ ﻋﻤﻮدﯾ ﺎً ﻋﻠ ﻰ وﺣﺪة‬
‫ﺄ‬
Ce = - 0.5
Ce = - 0.7
2
C e = + 0.8
where:
Pe =
+
+
Pe = C e k q
Ce = - 0.5
(0.00)
Ce = + 0.8
Suction
‫اﻟﻤﺴ ﺎﺣﺎت ﻋﻠ ﻰ اﻷﺳ ﻄﺢ اﻟﺨﺎرﺟﯿ ﺔ ﻟﻠﻤﻨﺸ‬
Ce = - 0.7
Ce =
‫( أو ﺳ ﺤﺐ‬Pressure) ‫ﻣﻌﺎﻣﻞ ﺗﻮزﯾ ﻊ ﺿ ﻐﻂ‬
Plan
‫( اﻟﺮﯾ ﺎح ﻋﻠ ﻰ اﻷﺳ ﻄﺢ اﻟﺨﺎرﺟﯿ ﺔ‬Suction)
C e = 0.8 + 0.5 = 1.3
k = Factor of exposure
Ground Roughness Length
Zone (A):
Zone (B):
Open exposure
Suburban exposure
Zone (C): City center exposure
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Page No.
Zone
Ground roughness
length
Height (m)
0 - 10
10 - 20
20 - 30
30 - 50
50 - 80
80 - 120
120 - 160
160 - 240
A
B
C
0.05
0.30
1.00
1.00
1.15
1.40
1.60
1.85
2.10
2.30
2.50
k
1.00
1.00
1.00
1.05
1.30
1.50
1.70
1.85
1.00
1.00
1.00
1.00
1.00
1.15
1.30
1.55
(Most critical case)
NOTE
(Open exposure or Suburban exposure or City center exposure)
(Ground roughness length = 0.05 or 0.30 or 1.00)
(Zone A)
k
(Most critical case)
q (kN/m )
2
q
2
0.5 r V C t Cs
=
1000
2
kN/m
where:
V (m/sec)
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Page No.
V (m/sec)
42
39
36
33
30
r=
1.25 kg/m 3
C t = Factor of topography
Ct
1.00
1.20
1.40
1.60
1.80
1.80
1.00
:
1.80
NOTE
1.00
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Ct
)
Page No.
C s = Structural factor
Turbulence
:
1.00
1.00
Cs
NOTE
C s & Ct
V
2
+
+
+
q = 0.5 r V C t Cs = 0.5 1.25 1.00 1.00 V 2
1000
1000
+
q = 6.25 10-4 V 2
q
(kN/m) 2
1.10
0.95
0.81
0.68
0.56
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Page No.
kN
kN
+
+
+ +
Wind load (F) = Pe area
= C e k q area
where:
area = Area subjected to wind
‫اﻟﻤﺴ ﺎﺣﺔ اﻟﻤﻌﺮﺿ ﺔ ﻟﻠﺮﯾ ﺎح‬
NOTE
‫ﺄﺛﯿﺮ ﺣﻤﻞ اﻟ ﺰﻟﺰال و اﻟﻌﻜ ﺲ و ذﻟ ﻚ ﻷن ﺣ ﺪوﺛﮭﻤﺎ‬
‫ﺎ ﻻ ﻧﺄﺧ ﺬ ﺗ‬
‫ﺄﺛﯿﺮ ﺣﻤﻞ اﻟﺮﯾ ﺎح ﻓﺈﻧﻨ‬
‫ﻋﻨ ﺪﻣﺎ ﻧﺄﺧ ﺬ ﺗ‬
ً‫ﻣﻌ ﺎً ﻧ ﺎدر ﺟ ﺪا‬.
Distribution of wind
+
+
Pe = C e k
q
2
kN/m
( P ) kN\m
(Most critical case) 4
Zone A
( P ) kN\m
3
k 4 = 1.60
q
( P2 ) kN\m
k 3= 1.40
q
k 2= 1.15
q
( P1 ) kN\m
q
k 1 = 1.00
2
2
+
+
k4
k3
k2
k1
+ +
+ +
+
+
P4 = Ce
P3 = Ce
P2 = Ce
P1 = Ce
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2
2
)
h4
h3
H
h2
h1
Page No.
+
+
+
F4
kN
+
(h4 b )
(h3 b )
+
F3 kN
(h2 b )
(h1 b )
H4
F2 kN
+
+
+
+
F4 = P4
F3 = P3
F2 = P2
F1 = P1
H3
+
+
b
q area
F= Ce k
F1
H2
kN
h
4
h
3
h
2
h
H1
H
1
+
+
+
+
Total Moment at base = S Fi H i
M overturning = F1 H 1 + F2 H 2 + F3 H 3 + ....................
NOTE
Working loads
Check sliding and Check overturning
Difference between wind & seismic loads:
Wind loads
Seismic loads
‫ﺔ ﻓ ﻲ اﺗﺠ ﺎه واﺣﺪ‬
‫ ﻗ ﻮة أﻓﻘﯿ‬-
‫ ﻗ ﻮة ﺗﮭ ﺰ اﻟﻤﻨﺸ ﺄ ﻓ ﻲ ﺟﻤﯿ ﻊ اﻻﺗﺠﺎھ ﺎت‬-
‫ ﺿ ﻐﻂ اﻟﺮﯾ ﺎح ﯾﻌﺘﻤ ﺪ ﻋﻠ ﻰ اﻟﺴ ﻄﺢ‬‫ﺄ‬
‫ ﻗ ﻮة اﻟ ﺰﻟﺰال ﻧﺴ ﺒﺔ ﻣﻦ وزن اﻟﻤﻨﺸ ﺄ‬-
‫اﻟﺨ ﺎرﺟﻲ ﻟﻠﻤﻨﺸ‬
‫ ﻗ ﻮة اﻟﺮﯾ ﺎح ﺗﺆﺛ ﺮ ﻋﻠ ﻰ اﻟﻮاﺟﮭﺔ ﺣ ﺘﻰ‬-
‫ ﻗ ﻮة اﻟ ﺰﻟﺰال ﺗﺆﺛ ﺮ ﻓ ﻲ ﻣﺴ ﺘﻮى اﻟﺒﻼﻃ ﺎت ﺣ ﺘﻰ‬-
‫ﻣﻨﺴ ﻮب ﺳ ﻄﺢ اﻷرض‬
Working loads ‫ﻮن‬
‫ أﺣﻤﺎل اﻟﺮﯾ ﺎح ﺗﻜ‬-
‫ﯿﺲ‬
Ultimate loads ‫ﻮن‬
‫ﻣﻨﺴ ﻮب اﻟﺘﺄﺳ‬
‫ أﺣﻤﺎل اﻟ ﺰﻻزل ﺗﻜ‬-
(0.00)
Wind load distribution
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Storey forces due to seismic load
)
Page No.
Factored loads of Ultimate Limit Design Method:
Ultimate Load (U)
:
0.8 [1.4 D.L. + 1.6 L.L. + 1.6 W.L.]
U=
1.12 D.L. + a L.L. + S.L.
1.4 D.L. + 1.6 L.L.
(U)
where:
D.L. = Dead load
L.L. = Live load
S.L. = Seismic load
W.L. = Wind load
a=
a
0.25
0.50
1.00
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Page No.
Example.
3.0 +16 = 48 m
The shown figure of a store house which lies at Cairo. The building
consist of a ground , mezzanine and 16 repeated floors and two
basements. It is required to :
1- Calculate the wind load acting on the building .
2- Check The Stability of the building against sliding and
overturning.
Given that :
3.5 3.5 4.0 4.0
t s average= 0.20 m
F.C. = 2.0 kN/m 2
Walls = 3.0 kN/m 2
L.L. = 3.0 kN/m 2
Sec . Elevation
6.0
6.0
6.0
6.0
6.0
6.0
6.0
6.0
6.0
6.0
Plan
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Page No.
Solution
Ce = 0.8 + 0.5 = 1.3
2
+ + +
+ +
+ + +
2
1.0 0.68 = 0.884 kN/m
1.15 0.68 = 1.017 kN/m2
1.40 0.68 = 1.238 kN/m2
1.60 0.68 = 1.414 kN/m2
1.85 0.68 = 1.635 kN/m 2
294.3 kN
1.635
2
kN/m
6m
+
+
+
F5 = P5 ( h b)
= 1.635 6 30 = 294.3 kN
+
+
+
+
+
+
+
q = 1.3
q = 1.3
q = 1.3
q = 1.3
q = 1.3
+ +
+
+
+
+
+
P1 = Ce
P2 = Ce
P3 = Ce
P4 = Ce
P5 = Ce
k1
k2
k3
k4
k5
+
+
+
-4
0.5 r V C t Cs
2
2
q=
= 6.25 10 33 = 0.68 kN/m
1000
Zone A ( More Critical )
Pe = Ce k q
(kN/m2 )
+
371.4 kN
1.2382
kN/m
10 m
1.017
kN/m2
10 m
56 m
20 m
+
+
+
F2 = P2 ( h b)
= 1.017 10 30 = 305.1 kN
+
+
+
F1 = P1 ( h b)
=0.884 10 30 = 265.2 kN
305.1 kN
265.2 kN
0.884
kN/m2 10 m
Total wind force = 265.2 + 305.1 + 371.4 + 848.4 + 294.3
= 2084.4 kN
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Page No.
7m
+
1.414
kN/m2
+
+
+
F3 = P3 ( h b)
=1.238 10 30 = 371.4 kN
+
848.4 kN
+
+
+
+
F4 = P4 ( h b)
= 1.414 20 30 = 848.4 kN
1.635
2
kN/m
848.4 kN
371.4 kN
60
47
305.1 kN
6m
1.414
kN/m2
20 m
1.2382
kN/m
10 m
1.017
kN/m2
10 m
56 m
294.3 kN
32
265.2 kN
0.884
kN/m2
22
10 m
7m
12
Check overturning
+
+ +
+
+
+
+
+
+
+
+
Total Moment at base ( Overturning Moment ) =
Fi H i
= F1 H1 + F2 H2 + F3 H3 + F4 H4 + F5 H5
= 265.2 12 + 305.1 22 + 371.4 32 + 848.4 47 + 294.3 60
= 79312.2 kN.m
+
Resisting Moment = WTotal
B
2
+
+
+
+
ws = t av gc + F.C. + Walls + L.L.
ws = 0.20 25 + 2.0 + 3.0 + 3.0 = 13.0 kN/m2
wFloor = 13.0 30 30 = 11700 kN
wTotal = 11700 20 = 234000 kN
+
Resisting Moment = 234000
30 3510000 kN.m
2 =
Resisting Moment
3510000
Factor Of Safety = Over Turning Moment = 79312.2 = 44.2 > 1.5
Safe
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Page No.
Check sliding
W = 0.30 234000 = 70200
+
Resisting Force =m
+
Sliding Force = wind force = 2084.4 kN
Resisting Force
70200
Factor Of Safety = Sliding Force = 2084.4
= 33.7 > 1.5
Safe
NOTE
Working loads
Check sliding and Check overturning
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Page No.
Systems resisting lateral loads.
‫ﺔ و ﻣﻨﮭﺎ‬
‫ﺎﺋﯿﺔ ﺗﻘ ﺎوم اﻟﻘ ﻮى اﻷﻓﻘﯿ‬
‫ﺗﻮﺟ ﺪ ﻋﺪة أﻧﻈﻤ ﺔ إﻧﺸ‬
1- Shear walls or cores
Shear walls
Core (Tube)
2- Coupled shear walls
3- Frames
Ductile frames (Beams + Columns)
4- Combination between different systems
- Frames + Shear walls
- Frames + Core
- Core + Shear walls
- Frames + Shear walls + Cores
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Core
Shear
wall
)
Frames
Page No.
Center of Mass (C.M.) & Center of Rigidity (C.R.)
( Columns - Core - Shear Walls )
( Slabs - Beams )
- ‫ﺔ ﻻﺑ ﺪ ﻣﻦ ﻣﻌﺮﻓ ﺔ ﺑﻌ ﺾ اﻟﻤﻔ ﺎھﯿﻢ‬
‫ﺎﺋﯿﺔ اﻟ ﺘﻲ ﺗﻘ ﺎوم اﻟﻘ ﻮى اﻷﻓﻘﯿ‬
‫ﻗﺒ ﻞ دراﺳ ﺔ اﻷﻧﻈﻤ ﺔ اﻹﻧﺸ‬
‫ﯿﺔ‬
Center of mass (C.M.):
‫اﻷﺳﺎﺳ‬
It is the center of gravity of area and it is the point of application
of ( Fb ).
wi Ai x i
wi A i
+
Datum
+
+
X C.M.=
where:
+
+ +
+ +
w2 A2
C.G. C.M.
Center of rigidity (C.R.):
It is the point where the force ( Fb ) is resisted.
( Stiffness )
( Columns - Core - Shear Walls )
NOTE
(C.M.)
( C.R.)
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Page No.
+
w1 A1
X C.M.
w A x + w2 A 2 x 2
X C.M.= 1 1 1
w1 A1 + w2 A2
+
x1
+
XC.M.= Distance between C.M. & datum
x i = Distance between C.G. of A i & datum
wi = Weight of floor at this part of floor A i
x2
C.G.
(C.R. )
(Shear wall )
(C.M.)
(Core)
( C.M.)
(C.R. )
(Torsion )
( C.R. )
Fb
Fb
(Torsion)
C.R.
C.M.
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C.R.
)
C.M.
Page No.
1- Shear walls or cores
Case (a): Symmetrical shear walls (C.M. = C.R.)
Lateral Force (F b )
Ii
+
Fi =
i=n
i=1
Ii
C.M.
‫ ﻋﻠ ﻰ اﻟﺤ ﻮاﺋﻂ ﺑﻨﺴ‬Bending Moment ‫ أو‬Shear force
Fb & M i =
Ii
i=n
i=1
+
( Inertia ) ‫ﺒﺔ‬
C.R.
Ii
M base
sw1
I1
I1
F1 =
F
2I1 + 2I 2 b
I2
F2 =
Fb
2I1 + 2I 2
where:
sw2
sw2
I2
I2
C.M.
sw1
I1
+
C.R.
x2
x1
+
x1
x2
Fb
F b = Force acting at the base
Fi = Force acting on the shear wall no. ( i )
I i = Moment of inertia of the shear wall no. ( i )
M base= Bending Moment acting at the base
M i = Bending Moment acting on the shear wall no. ( i )
Case (b): Unsymmetrical shear walls in one direction (C.M. = C.R.)
C.R.
C.M.
L
To Get C.R.
Datum
1
X
C.R.
Ii xi
Ii
=
sw2
sw1
+
+
+
+
I1
x1
I3
e
xC.M.
x2
x3
e = Distance between C.M. & C.R.
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sw4
C.M.
I2
xC.R.
e = X C.M. X C.R.
where:
sw3
C.R.
I1 x 1+ I 2 x 2 + I 3 x 3 + I 4 x 4
X C.R.=
I1 + I2 + I 3 + I 4
2
MT
Fb
x4
Page No.
I4
M T = Fb e*
e *= e + 0.05 L
+
3
‫ ( ﻋﻠ ﻰ اﻟﺤ ﻮاﺋﻂ ﻛ ﺎﻵﺗﻲ‬F ) ‫ﺛ ﻢ ﺗ ﻮزع اﻟﻘ ﻮة‬
b
Ii xi
MT
i=n
2
I (xi )
i=1 i
+
+
Fi = i = In i Fb +
Ii
i=1
where:
Fb
x = (C.R.) ‫ھﻮ ﺑﻌ ﺪ اﻟﺤ ﺎﺋﻂ ﻋﻦ‬
i
sw3
C.M.
x1
I3
e
x3
x2
+
Ii
Ii
i=1
i=n
x4
Fb
+
Ii xi
2
I
(
x
)
i
i
i=1
i=n
I4
MT
I1
I1 x 1
Fb 2
2
2
2
I1 + I2 + I3 + I4
I 1( x1) + I2 (x2) + I 3 (x3) + I 4 (x4)
F2 =
I2
I2 x 2
Fb 2
2
2
2
I1 + I2 + I3 + I4
I 1( x1) + I2 (x2) + I 3 (x3) + I 4 (x4)
F3 =
I3
I3 x 3
Fb +
2
2
2
2
I1 + I2 + I3 + I4
I 1( x1) + I2 (x2) + I 3 (x3) + I 4 (x4)
F4 =
I4
I4 x 4
Fb +
2
2
2
2
I1 + I2 + I3 + I4
I 1( x1) + I2 (x2) + I 3 (x3) + I 4 (x4)
+
+
+
+
F1 =
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Page No.
+
I2
MT
+
C.R.
I1
sw4
MT
+
sw1
MT
MT
+
sw2
MT
Fb
NOTE
sw2
sw1
MT
sw4
sw3
C.M.
C.R.
(- ve)
Zone ( 2 )
(+ ve)
Zone ( 1 )
C.M. ‫ و اﻟ ﺘﻰ ﺑﮭ ﺎ‬Zone ( 1 ) ‫ إذا ﻛ ﺎن اﻟﺤ ﺎﺋﻂ ﯾﻘ ﻊ ﻓ ﻲ‬+ ve - ‫ﺣﯿ ﺚ أن اﻹﺷ ﺎرة ﺗﻜ ﻮن‬
.Zone ( 2 ) ‫ إذا ﻛ ﺎن اﻟﺤ ﺎﺋﻂ ﯾﻘ ﻊ ﻓ ﻲ‬- ve ‫و ﺗﻜ ﻮن اﻹﺷ ﺎرة‬
.C.R.
‫ ﯾﻜ ﻮن اﻟﺨ ﻂ اﻟﻤ ﺎر ﺑـ‬Zone ( 1 ) & Zone ( 2 ) - ‫ﻞ ﺑﯿ ﻦ‬
‫اﻟﺨ ﻂ اﻟﻔﺎﺻ‬
(Shear wall) ‫ ( ﯾﻤﻜ ﻦ إﯾﺠ ﺎد ﻧﺴ ﺒﺔ اﻟﻘ ﻮى اﻟﻤﺆﺛ ﺮة ﻋﻠ ﻰ ﻛ ﻞ‬F = 1 kN ) - ‫ﺎﻟﺘﻌﻮﯾﺾ ﻋﻦ‬
b
+
M i = % of each Shear Wall
+
Ii xi
(1
2
I (x i)
i=1 i
i=n
+
+
% of each Shear Wall = i = In i 1 +
Ii
i=1
‫ﺑ‬
e* )
M base
Case (c): Unsymmetrical shear walls in both direction
(ey ) & (ex )
( Eccentricity )
( C.R.)
( I y ) & ( Ix )
(I)
y
Lx
sw3
Fby e *x
+
sw1
xC.M.
sw4
xC.R.
Fbx
Ly
C.M.
Fbx
+
x1
e *y
ey C.R.
ex
sw2
y2
x4
y3
yC.R. yC.M.
x
F by
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Page No.
e x* = e + 0.05 L x
1
2
(sw3 ) & (sw2)
Y = Iy i yi = I y2 y2 + Iy3 y3
C.R.
Iy i
Iy2 + Iy3
e y = Y C.M. - Y C.R.
3
e y* = e + 0.05 L y
( Fby )
+
3
+
M T y = Fby e*x
( Fbx)
+
+
M T x = Fbx e*y
+
1
+
2
(sw4 ) & (sw1)
X = Ix i xi = Ix1 x 1 + Ix 4 x 4
C.R.
Ix i
Ix1+ Ix4
e x = X C.M. - X C.R.
y
Lx
sw3
sw1
x1
Fbx
ey
sw 2
i=n
i=1
i=n
x4
C.M.
sw4
y
Ly
3
C.R.
ex
y2
2
2
2
2
2
2
x
Ix i (x i) = I x1 (x1) + Ix 4 (x4)
Fby
Iy i (y i) = I y2 (y2 ) + Ix 3 (y3 )
i=1
M Ty
Iy i yi
2
2
[
I
(
x
)
+
I
(
y
)
y
x
i i
i i
i=1
M Tx
i=n
+
( C.M. )
( C.M. )
+
Fx = i = In yi Fbx +
i
Iy i
i=1
+
i=n
+
Ix i xi
2
2
[
I
(
x
)
+
I
(
y
)
x
y
i
i
i
i
i=1
Fy = i = In xi Fby +
i
Ix i
i=1
( Shear wall )
( Shear wall )
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(+ )
( )
)
Page No.
NOTE
sw1
Symmetric
e *= 0.05 L
M T = Fb e*
sw2
sw2
I1
I2
I2
D.L.
T.L.
T.L.
sw1
+
e min = 0.05 L
sw1
I1
I1
sw1
sw1
I1
I1
I2
C.R.
e min
I2
x1
I2
sw1
I1
emin
C.M.
= 0.05L
x2
x1
x1
x2
x2
x1
Fb
Ii xi
i=n
2
I
(
x
)
i
i=1 i
+
+
sw2
C.R.
Fb
Fi = i = In i Fb +
Ii
i=1
I1
C.M.
I2
OR
C.M.
sw1
D.L.
sw2
= 0.05L
x2
I2
T.L.
sw2
I2
sw2
sw2
T.L.
C.M.
sw2
I1
sw1
MT
‫ ( ﻋﻠ ﻰ اﻟﺤ ﻮاﺋﻂ ﻛ ﺎﻵﺗﻲ‬F ) ‫ﺛ ﻢ ﺗ ﻮزع اﻟﻘ ﻮة‬
b
where:
x = (C.R.) ‫ھﻮ ﺑﻌ ﺪ اﻟﺤ ﺎﺋﻂ ﻋﻦ‬
i
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( 0.05 L Fb )
+
Fb +
( 0.05 L Fb )
+
I2
2 ( I1 + I2 )
+
F2 =
Fb +
More Critical
I1 x 1
2
2
2 [ I 1( x1) + I 2 (x2) ]
I2 x 2
2
2
2 [ I 1( x1) + I 2 (x2) ]
+
I1
2 ( I1 + I2 )
+
F1 =
+
e min
Page No.
Illustrative Example
For the shown figure , if the thickness of all shear walls = 25 cm .
It is required to :
1- Calculate the percentage of force ( P ) acting on each wall
Solution
3
3.5
2.5
sw1
sw2
+
Isw1 = 0.25 (3.5) = 0.8932 m4
12
sw3
4.0
3
+
I sw2 = 0.25 (2.5) = 0.3255 m4
12
Fb
15
3
+
I sw 3 = 4.0 ( 0.25 ) = 0.0052 m4
12
sw3
i=6
i=6
i=1
I
I sw 2
i=6
i=1
I
Isw 3
i=6
i=1
I
sw2
sw1
I = 2 ( 0.8932 + 0.3255 + 0.0052 )
i=1
4
= 2.4478 m
Isw 1
4.0
3.5
10.0 m
0.8932
= 2.4478
= 36.50 %
0.3255
= 2.4478
= 13.3 %
0.0052
= 0.20 %
= 2.4478
( Inertia )
( sw3 )
NOTE
Shear wall
Fb
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Page No.
2.5
Design of Shear wall.
Shear walls solved as a cantilever totally fixed in foundation and
subjected to moment (M ) due to lateral force and normal force
( N ) due to own weight ,weight of walls and floors .
2
sw
F6
F5
F4
1
sw
sw 3
sw 6
sw 4
F6
F5
sw 5
F3
F2
F1
F4
F3
Elevation
F2
F1
sw1
sw2
sw3
Fi
F 6-2
sw4
F 5-2
sw6
sw5
F 4-2
Plan
F 2-2
Fsw =
F 1-2
i
i=n
i=1
Shear Wall 2
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I sw i
)
+
F 3-2
Fi
I sw i
Page No.
= % of each Shear Wall
+
+
MS.L.= % of each Shear Wall M
Base Seismic U.L.
N
load
+
+
M W.L.= % of each Shear Wall M Base Wind Working = % of each Shear Wall
Fi H i
Fi H i
= O.W of shear wall + The Normal Force Due to Seismic or Wind Loads
+
+ The part from the floor which it carry
number of floors
Cases of Load Combinations
Case (1)
D.L.+L.L.
Case (2)
D.L.+L.L.+W.L.
Case (3)
D.L.+L.L.+S.L.
N = 1.4 N D.L.+ 1.6 N L.L.
M = 1.4 M D.L.+ 1.6 M L.L.
N = 0.8 [1.4 ND.L.+ 1.6 N L.L.+ 1.6 NW.L.]
M = 0.8 [1.4 MD.L.+ 1.6 M L.L.+ 1.6 M W.L.]
N = 1.12 ND.L.+ a N L.L.+ N S.L.
M = 1.12 M D.L.+ a M L.L.+ M S.L.
where:
a=
a
0.25
0.50
1.00
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Page No.
NOTE
Working
Manual
MW.L.
Ultimate
MS.L.
Seismic & Wind
SAP
N W.L.= Zero
and
N S.L.= Zero
D.L. & L.L.
Flat Slabs
Moments
Frames
Cases of Load Combinations
Case (1) N = 1.4 N D.L.+ 1.6 N L.L.
Case (2) N = 0.8 [1.4 ND.L. + 1.6 N L.L.] & M = 0.8 [1.6 M W.L.]
Case (3) N = 1.12 ND.L.+ a N L.L.
Design of Shear wall under M & N
e
t
<
1
0.5
+
Msu = N u.l.
0.225
J
Get
0.15 b d
100
st. 360/520
N u.l.
=
Fcu b t
Fcu b t 2
As= As = (
\
Get
=
-4
cu
F
0.6
As =
Ac
min
100
As min= 1.3A s req
st. 240/350
e < 0.5
t
M u.l.
Fcu
b d > 1.1 b d
Fy
Fy
0.25 b d
100
Mu.l.
N u.l.
Using interaction diagram
As = As\
take x = 0.9
+
+
+
e=
2
e s = e + 2t - c
es
Mus
C1
Fcu b
M us
Nu.l.
As =
J d Fy ( Fy / s )
d = C1
M = M S.L.
&
As
10 ) b t
As
b
t
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Page No.
2- Rigid frames
frames ‫ﺾ ﻟﻜ ﻲ ﺗﻜ ﻮن‬
‫ھﻲ ﻋﺒ ﺎرة ﻋﻦ ﻣﺠﻤﻮﻋﺔ ﻣﻦ اﻟﻜﻤ ﺮات و اﻷﻋﻤ ﺪة ﻣﺘﺼ ﻠﺔ ﺑﺒﻌ‬
‫ ﻟﻤﻘﺎوﻣ ﺔ اﻟﻘ ﻮى اﻷﻓﻘﯿ‬frames ‫ﺘﺨﺪم ھﺬه‬
‫ﺔ و ﻓ ﻲ ھﺬه اﻟﺤﺎﻟ ﺔ ﯾﺠ ﺐ أن ﺗﺼ ﻤﻢ اﻟﻮﺻ ﻠﺔ‬
‫ﺔ‬
‫ﺘﻄﯿﻊ ﻣﻘﺎوﻣﺔ اﻷﺣﻤ ﺎل اﻷﻓﻘﯿ‬
‫و ﺗﺴ‬
‫ﻟﻜ ﻲ ﺗﺴ‬. (rigid frames) ‫ﺑﯿ ﻦ اﻟﻌﻤ ﻮد و اﻟﻜﻤ ﺮة ﻋﻠ ﻰ أﻧﮭ ﺎ‬
point of
zero
moment
B.M.D due to Hz. loads
Rigid Frames
Rigid Frames
Fb
Plan
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Page No.
‫‪NOTE‬‬
‫ﯾﺠ ﺐ أن ﺗﻜ ﻮن أﻃﻮال ﺟﻤﯿ ﻊ وﺻ ﻼت اﻟﺤﺪﯾ ﺪ )‬
‫‪.‬ﺗﺤﻤ ﻞ اﻟﻌ ﺰوم اﻟﻨﺎﺗﺠ ﺔ ﻋﻦ اﻟﻘ ﻮى اﻷﻓﻘﯿ‬
‫‪ (L d= 60‬أي وﺻ ﻼت ﺷ ﺪ ﻟﻜ ﻲ ﺗﺴ‬
‫ﺘﻄﯿﻊ‬
‫ﺔ‬
‫ﯾﺠ ﺐ ﻋﻤﻞ ﺟﻤﯿ ﻊ وﺻ ﻼت اﻟﺤﺪﯾ ﺪ ﻟﻸﻋﻤ ﺪة ﻋﻨ ﺪ ﻣﻨﺘﺼ ﻒ اﻟ ﺪور )‪(point of zero moment‬‬
‫ﻧﻈ ﺎم )‪ (rigid frames‬ﯾﺼ ﻠﺢ ﻟﻤﻘﺎوﻣ ﺔ اﻷﺣﻤ ﺎل ﺣ ﺘﻰ‬
‫اﻷﻓﻘﯿ‬
‫ﺔ ﻛﺒ‬
‫ﯿﺮة ﻧﺴ‬
‫دور و ﻟﻜ ﻦ ﺗﻜ ﻮن اﻹزاﺣﺔ‬
‫ﺒﯿﺎً ﻟ ﺬﻟﻚ ﯾﺠ ﺐ ﺣﺴ ﺎﺑﮭﺎ‬
‫وﺻ ﻼت اﻟﺤﺪﯾ ﺪ ﻟﻸﻋﻤ ﺪة‬
‫ﻋﻨ ﺪ ﻣﻨﺘﺼ ﻒ اﻻرﺗﻔ ﺎع‬
‫‪Ld = 60‬‬
‫‪Ld = 60‬‬
‫‪Ld = 60‬‬
‫‪Ld = 60‬‬
‫‪Details of RFT of a rigid joint‬‬
‫ﻋﻨ ﺪ اﺳ ﺘﺨﺪام ‪Core & Shear walls‬ﻛﺎﻧ ﺖ اﻟﻘ ﻮة اﻷﻓﻘﯿ‬
‫ﺔ ) ‪Base shear ( Fb‬‬
‫و اﻟﻌ ﺰوم اﻟﻨﺎﺗﺠ ﺔ ﻋﻨﮭ ﺎ‪ M base‬ﺗ ﻮزع ﻋﻠ ﻰ اﻟﻌﻨﺎﺻ ﺮ اﻟﻤﻘﺎوﻣ ﺔ ﺑﺪﻻﻟ ﺔ ) ‪Inertia ( I‬‬
‫ﺑﯿﻨﻤ‬
‫ﺎ ﻋﻨ ﺪ اﺳ ﺘﺨﺪام ‪ Rigid frames‬ﺗ ﻮزع اﻟﻘ ﻮة اﻷﻓﻘﯿ‬
‫ﺔ )‪Base shear ( Fb‬‬
‫و اﻟﻌ ﺰوم اﻟﻨﺎﺗﺠ ﺔ ﻋﻨﮭ ﺎ‪ M base‬ﻋﻠ ﻰ اﻟﻌﻨﺎﺻ ﺮ اﻟﻤﻘﺎوﻣ ﺔ ﺑﺪﻻﻟ ﺔ ) ‪Stiffness ( K‬‬
‫‪Page No.‬‬
‫)‬
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Case (a): Symmetrical rigid frames (C.M. = C.R.)
Lateral Force (Fb )
(stiffness of each frame) ‫ﺒﺔ‬
C.R.
C.M.
‫( ﺑﻨﺴ‬Frames) ‫ ﻋﻠ ﻰ‬Bending momen ‫ أو‬Shear force
To get stiffness of each frame:
Fb
For typical story
1
2I c1
1+
h ( Ibb1 + Ibb2 )
1
2
12EI
K int.= 3 c1
h
K ext.=
1
2I c2
h ( Ibb1 )
1
12EI c2
1+
h3
F1
F2
h =
F1
K2
K1
C.M.
C.R.
K1
K2
h Ic2
Ic1
Ib1
h Ic2
where:
F2
Ib2
Ic2
Ic1
b1
‫ارﺗﻔ ﺎع اﻟ ﺪور‬
b2
E = 4400 Fcu
K = SK
Fi
+int.
SK
K ext.
ext.
K int.
K ext.
KF1= 2Kext.+ Kint. ( stiffness for frame F1 )
KF2= 2Kext.+ 2K int. ( stiffness for frame F2 )
K Fi
i=n
i=1
K Fi
Fb
&
‫( ﺑﻨﺴ‬Frames) ‫ ( ﻋﻠ ﻰ‬F ) ‫ﺛ ﻢ ﺗ ﻮزع اﻟﻘ ﻮة‬
b
Fi
M i = i =K
n
i=1
+
Fi =
+
(Stiffness of each frame) ‫ﺒﺔ‬
KFi
M base
‫ﺘﻨﺘﺠﺔ ﻋﻠ ﻰ أﺳ ﺎس ﺛﺒ ﺎت ﻗﻄ ﺎع اﻟﻌﻤ ﻮد ﻓ ﻲ ﻛ ﻞ اﻷدوار‬
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‫ﺎﺑﻘﺔ ﻣﺴ‬
Ic2
‫اﻟﻤﻌ ﺎدﻻت اﻟﺴ‬.
Page No.
+
Case (b): Unsymmetrical rigid frames in one direction (C.M. = C.R.)
C.R.
C.M.
L
Datum
To Get C.R.
F1
F2
F3
F4
x
1 XC.R. = K Fi i
MT
K Fi
C.M.
C.R.
2
+
+
+
K F1 x1 + KF2 x 2 + KF3 x 3 + K F4 x4
K F1 + KF2 + K F3 + K F4
+
X C.R.=
e = X C.M. - X C.R.
K1
K2
K3
xC.R.
e
xC.M.
x1
where:
x2
x3
e = Distance between C.M. & C.R.
x4
M T = Fb e*
e * = e + 0.05 L
+
3
Fb
+
KFi xi
i=n
2
K
(
x
)
i
Fi
i=1
+
Fi
Fi = i = K
Fb +
n
K Fi
i=1
+
‫ ( ﻛ ﺎﻵﺗﻲ‬Frames ) ‫ ( ﻋﻠ ﻰ‬F ) ‫ﺛ ﻢ ﺗ ﻮزع اﻟﻘ ﻮة‬
b
MT
where:
x = (C.R.) ‫( ﻋﻦ‬frames)
‫ھﻮ ﺑﻌ ﺪ‬
i
NOTE
F1
F2
Fb
MT
C.R.
(- ve)
Zone ( 2 )
F3
F4
C.M.
(+ ve)
Zone ( 1 )
C.M. ‫ و اﻟ ﺘﻰ ﺑﮭ ﺎ‬Zone ( 1 ) ‫ ﯾﻘ ﻊ ﻓ ﻲ‬frame ‫ إذا ﻛ ﺎن‬+ ve - ‫ﺣﯿ ﺚ أن اﻹﺷ ﺎرة ﺗﻜ ﻮن‬
.Zone ( 2 ) ‫ﯾﻘ ﻊ ﻓ ﻲ‬frame ‫ إذا ﻛ ﺎن‬- ve ‫و ﺗﻜ ﻮن اﻹﺷ ﺎرة‬
.C.R. ‫ ﯾﻜ ﻮن اﻟﺨ ﻂ اﻟﻤ ﺎر ﺑـ‬Zone ( 1 ) & Zone ( 2 ) - ‫اﻟﺨ ﻂ اﻟﻔﺎﺻ ﻞ ﺑﯿ ﻦ‬
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Page No.
K4
(Frame) ‫( ﯾﻤﻜ ﻦ إﯾﺠ ﺎد ﻧﺴ ﺒﺔ اﻟﻘ ﻮى اﻟﻤﺆﺛ ﺮة ﻋﻠ ﻰ ﻛ ﻞ‬F = 1kN) - ‫ﺎﻟﺘﻌﻮﯾﺾ ﻋﻦ‬
M i = % of each Frame
+
+
(1
‫ﺑ‬
e* )
M base
‫ﺎﺑﻘﺔ ﺗﻜ ﻮن ﻟ ﻼدوار اﻟﻤﺘﻜ ﺮرة أﻣﺎ ﻟﻠ ﺪور اﻷول‬
‫ اﻟﻤﺤﺴ ﻮﺑﺔ ﻟﻠ ﺪور‬stiffness (K) ‫ﺘﺨﺪم‬
KFi xi
2
K
(
x
)
i
Fi
i=1
i=n
+
1 +
+
Fi
% of each Frame = i = K
n
K Fi
i=1
+
b
‫ﻮاﻧﯿﻦ اﻟﺴ‬
‫ﮭﯿﻞ ﻧﺴ‬
‫ اﻟﻤﺤﺴ ﻮﺑﺔ ﻣﻦ اﻟﻘ‬stiffness (K) -
‫و اﻷﺧ ﯿﺮ ﺗﻜ ﻮن ﻟﮭ ﺎ ﻗﯿ ﻢ آﺧﺮى و ﻟﻜ ﻦ ﻟﻠﺘﺴ‬
.(typical story)‫ﺮر‬
‫اﻟﻤﺘﻜ‬
For top story
24EI
K int.= 3 c1
h
I b1
1
h Ic2
3I c1
2+
h ( Ibb1 + Ibb2 )
1
2
24EI
K ext.= 3 c2 2 +
h
I b2
Ic2
Ic1
b1
1
3I c2
h ( Ibb1 )
1
b2
‫ﻟﻠﻘ ﺮاءة ﻓﻘ ﻂ‬
For bottom story
24EI
K int.= 3 c1
h
K ext.=
1
h Ic2
3I c1
2+
I
h ( Ibb1 + Ibb2 ) h c2
1
2
24EI c2
2+
h3
- If I b >>> I c
Ib1
Ic1
Ib2
Ic2
Ic1
b1
b2
1
3I c2
h ( Ibb1 )
1
The column considered as a fixed column
K=
12EI c
h3
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Ic2
Page No.
NOTE
Symmetric
e *= 0.05 L
M T = Fb e*
+
e min = 0.05 L
F1
F2
F2
F1
F1
F2
F2
F1
K1
K2
K2
K1
K1
K2
K2
K1
D.L.
T.L.
T.L.
T.L.
C.M.
D.L.
T.L.
C.M.
F1
F2
F2
F1
F1
F2
F2
F1
K1
K2
K2
K1
K1
K2
K2
K1
C.R.
emin
x1
x2
C.R.
OR
C.M.
C.M.
= 0.05L
x2
x1
x1
Fb
e
min
= 0.05L
x2
x2
x1
Fb
Fb +
KFi xi
i=n
2
K
(
x
)
i
Fi
i=1
+
+
Fi
Fi = i = K
n
K Fi
i=1
+
‫ ( ﻛ ﺎﻵﺗﻲ‬frames ) ‫ ( ﻋﻠ ﻰ‬F b) ‫ﺛ ﻢ ﺗ ﻮزع اﻟﻘ ﻮة‬
MT
where:
x i= (C.R.) ‫( ﻋﻦ‬frames)
‫ھﻮ ﺑﻌ ﺪ‬
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( 0.05 L Fb )
+
( 0.05 L Fb )
+
K2
2 ( K 1 + K2 )
+
F2 =
More Critical
K1 x1
Fb +
2
2
2 [ K1( x 1) + K 2 (x 2) ]
K2 x2
Fb +
2
2
2 [ K1( x 1) + K 2 (x 2) ]
+
K1
2 ( K 1 + K2 )
+
F1 =
+
e min
Page No.
Drift of structures due to seismic loads
Importance of drift:
Is to satisfy serviceability requirements.
The drift should be limited to fulfill the safety requirements for
non-structural elements.
‫ﻐﯿﻞ‬
‫ﺎﺋﯿﺔ (واﺟﮭﺎت‬
‫ﻢ اﻟﻤﺴ ﻤﻮح ﺑﮭ ﺎ ﻃﺒﻘ ﺎً ﻟﺤ ﺪود اﻟﺘﺸ‬
‫ﺎﻓﯿﺔ ﻋﻠ ﻰ اﻟﻌﻨﺎﺻ ﺮ اﻟﻐ ﯿﺮ إﻧﺸ‬
‫ﺄ ﻋﻦ اﻟﻘﯿ‬
‫ﺔ اﻟﺤﺎدﺛ ﺔ ﻟﻠﻤﻨﺸ‬
‫ﺒﺐ ﻓ ﻲ ﺗﻮﻟ ﺪ أﺣﻤﺎل إﺿ‬
‫ﯾﺠ ﺐ أﻻ ﺗﺰﯾ ﺪ اﻹزاﺣ ﺔ اﻟﻜﻠﯿ‬
‫اﻟﻤﻄﻠﻮﺑ ﺔ و ذﻟ ﻚ ﻟﻜ ﻲ ﻻ ﺗﺘﺴ‬
‫ ﺗ ﺆدي إﻟ ﻰ ﺗﺸ ﺮﺧﮭﺎ و اﻧﮭﯿﺎرھ ﺎ‬................) - ‫ ﺣ ﻮاﺋﻂ‬- ‫زﺟﺎج‬
Total Drift = Web Drift + Chord Drift
1- Web drift
Shear ‫ﺔ‬
‫ﺄ ﻧﺘﯿﺠ‬
‫ﺔ ﺗﺤ ﺪث ﻟﻠﻤﻨﺸ‬
‫ھﻲ إزاﺣﺔ أﻓﻘﯿ‬
web drift
D8
Q8
Q7
D7
Q6
D6
D5
Q5
Q4
D4
Q3
D3
Q2
Q1
Load diagram
Shear diagram
D2
D1
Drift diagram
Qi
+
+
% of each frame
K Fi
Total web drift for each frame ( SD i ) = S Q i % of each frame
K Fi
Web drift for each frame ( D i ) =
where:
Di = Story drift of each frame
Q i = Shear force acting on the story
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Page No.
KFi xi
2
K
(
x
)
i
Fi
i=1
i=n
(1
+
K Fi 1 +
K Fi
i=1
i=n
+
(For Symmetrical Rigid Frames)
+
% of each Frame =
K Fi
K Fi
i=1
i=n
+
% of each Frame =
e* )
(
For Unsymmetrical
Rigid Frames
Case (a): Web drift for symmetrical rigid frames
+
Web drift for each frame ( D i ) =
Qi
+
Qi
=
% of each frame
K Fi
K Fi
K Fi
i=1
i=n
=
K Fi
Qi
i=n
i=1
as K s =
i=n
i=1
K Fi
K Fi
where K s = Story stiffness
Q
Web drift ( D i) = Ki
s
Total web drift ( SD i) =
‫ﺎوﯾﺔ‬
D1
S Qi
‫ ﻣﺘﺴ‬frames ‫ﺔ ﻟﻜ ﻞ‬
D2
Ks
‫ﺔ ﺗﻜ ﻮن اﻹزاﺣ ﺔ اﻷﻓﻘﯿ‬
D2
‫ﻓ ﻲ ﺣﺎﻟ ﺔ اﻟﻤﻨﺸ ﺄت اﻟﻤﺘﻤﺎﺛﻠ‬
D1
Fb
F1
F2
F2
F1
C.M.
C.R.
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Page No.
)
Case (b): Web drift for unsymmetrical rigid frames
Qi
+
+
% of each frame
K Fi
Total web drift for each frame ( SD i ) = S Q i % of each frame
K Fi
Web drift for each frame ( D i ) =
‫ ﻋﻼﻗ ﺔ‬frames ‫ﺔ ﻟﻜ ﻞ‬
KFi xi
2
K
(
x
)
i
Fi
i=1
i=n
( 1 e* )
+
i=n
+
K Fi 1 +
K Fi
i=1
+
% of each Frame =
+
where:
‫ﺔ ﺗﻜ ﻮن اﻟﻌﻼﻗ ﺔ ﺑﯿ ﻦ اﻹزاﺣﺔ اﻷﻓﻘﯿ‬
( For Unsymmetrical
rigid frames )
‫ﻓ ﻲ ﺣﺎﻟ ﺔ اﻟﻤﻨﺸ ﺄت اﻟﻐ ﯿﺮ ﻣﺘﻤﺎﺛﻠ‬
‫ﺧﻄﯿ ﺔ‬
D1
D2
D3
straight line
F1
F2
D4
F3
F4
MT
C.R.
C.M.
Fb
NOTE
H
ww KN/m
‫ ﻓﺄﻧ ﮫ ﯾﻤﻜ ﻦ‬wind load ( w ) ‫ ( ﻣﺜ ﻞ‬uniform load ) ‫ﻓ ﻲ ﺣﺎﻟ ﺔ ﺗﻌ ﺮض اﻟﻤﻨﺸ ﺄ إﻟ ﻰ‬
w
‫ ( ﻣﻦ اﻟﻤﻌﺎدﻟ ﺔ‬web drift ) ‫ﺣﺴ ﺎب‬
h
y(x)
x
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2
y (x) = ww ( x H - 2x )
h Ks
)
Page No.
2- Chord drift
Bending ‫ﺔ‬
‫ﺄ ﻧﺘﯿﺠ‬
‫ﺔ ﺗﺤ ﺪث ﻟﻠﻤﻨﺸ‬
‫ھﻲ إزاﺣﺔ أﻓﻘﯿ‬
H
we KN/m
Chord drift
Mbase
Moment Diagram
B
we H
Chord drift (D c) = 8EI
e
Drift diagram
4
where:
I e = Composite moment of inertia of columns at C.G.
C.G
L2
L1
A1
A2
I1
I2
2
2
+
+
Ie = 2 [ I1 + A1 ( L1 ) ] + 2 [ I2 + A2 ( L 2 ) ]
+
M Fi = % of each frame
M base = we H
2
2
get we
NOTE
‫ ﻟ ﺬﻟﻚ‬web drift ‫ﺒﺔ ﻟﻘﯿﻤ ﺔ‬
‫ ﻗﯿﻤ ﺔ ﺻ ﻐﯿﺮة ﺟ ﺪاً ﺑﺎﻟﻨﺴ‬chord drift ‫ﻏﺎﻟﺒ ﺎً ﻣﺎ ﺗﻜ ﻮن ﻗﯿﻤ ﺔ‬
‫ﯾﻤﻜ ﻦ إھﻤﺎﻟﮭﺎ‬
DTotal = D web + D chord > D allowable =
H
500
600
Serviceability Requirements
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Page No.
Example.
For the shown figure of a residential building lies at Aswan . The
building consist of ( 8 ) repeated floors . L. L. = 2.0 kN/m 2&
2
F.C. = 1.5 kN/m and the soil is loose sand.
Due to Earthquake loads , it is required to :
1- Calculate the story shear at each floor level and draw its
distribution on the height of the building.
2- Find the bending moment and shearing forces acting at base
level of the R.C. columns and draw distribution of shear force
and bending moment.
3- Check Stability of the building against over turning
Given that :
+
4 + 8=32 m
+
All beams = 25 60
All columns = 25 80
Slab thickness = 160 mm
Elevation
7 +3 = 21 m
+
7 3 = 21 m
Plan
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Page No.
Solution
1- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
Fig. (8-1)
P. (1/5 code)
Aswan
Loose sand
Table (8-2)
P. (2/5 code)
Zone (2)
Table (8-1)
P. (2/5 code)
a g = 0.125 g
Soil type (D)
Soil type (D)
Table (8-3)
P. (3/5 code)
Response spectrum curve Type (1)
T1 = Ct H
S =
TB =
TC =
TD =
3/4
Beams + Columns
R.C. Frames
P. (5/5 code)
C t = 0.075
Total Height of building (H) = 32 m
3/4
+
T1 = 0.075 32
Check
Sd (T)
= 1.01sec.
T1 < ( 4 T C = 1.2 sec.)
O.K.
T1 < 2.0 sec.
O.K.
Sd (T1 )
Time (Sec)
TB TC
1.80
0.10
0.30
1.20
T1
TD
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4.0
)
Page No.
Eq. (8-13)
P. (3/5 code)
TC h
> 0.20 a g g1
T1
Table (8-9)
P. (4/5 code)
+
+
Sd (T1) = 0.125 g 1.00
g1 = 1.00
1.80
2.5
5.00
0.30
1.01
+
Sd (T1) = a g g1 S 2.5
R
R = 5.00
h = 1.00
+
TC < T1 < TD
1.00
= 0.0334 g
0.20 a g g1 = 0.20
+
+
0.125 g 1.00 = 0.025 g < Sd (T1)
O.K.
T1 > 2 TC = 0.60 sec
l = 1.00
P. (5/5 code)
+
+
ws = D.L. + a L.L.
ws = ( t s gc + F.C. + Walls ) + a L.L.
Table (8-7)
P. (5/5 code)
a = 0.25
+
+
+
+
+
+
+
+
+
+
+
+
ws = ( 0.16 25 + 1.5 ) + 0.25 2.0 = 6.0 kN/m2
beams
columns
slab
wFloor = 6.0 21 21 + 0.25 0.6 25 ( 21 8) + 0.25 0.8 4 25 (16)
= 3596 kN
+
wTotal = 3596 8 = 28768 kN
NOTE
ts
t av
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Page No.
l W
g
+
= 0.0334 g
1.0
+
+
Fb = Sd (T1)
28768
g
F b = 960.85 kN
2- Distribution of lateral force on each floor
+
Fi = Fb
w i Hi
w i Hi
i=1
i=n
w i Hi
H
= i=n i
w i Hi i = 1 Hi
i=1
i=n
‫ﻷن وزن اﻟ ﺪور ﺛﺎﺑ ﺖ‬
i=8
H = 32 + 28 + 24 + 20 +16 + 12 + 8 + 4
i
i=1
= 144 m
Fi = ( 960.85 ) 1 H i
144
28
20
Fi
=
16
6.6726 H i
12
4
Floor H (m) F (kN) Q (kN)
i
i
i
No.
32.0 213.52 213.52
8
8
Mi (kN.m)
854.09
7
28.0
186.83
400.35
2455.51
6
24.0
160.14
560.50
4697.49
5
20.0
133.45
693.95
7473.28
4
16.0
106.76
800.71
10676.11
3
12.0
80.07
880.78
14199.23
2
8.0
53.38
934.16
17935.87
1
4.0
26.69
960.85
21779.27
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Page No.
24
32
213.52
213.52
400.35
186.83
854.09
560.50
160.14
2455.51
693.95
133.45
800.71
106.76
26.69
7473.28
880.78
80.07
53.38
4697.49
10676.11
934.16
14199.23
960.85
17935.87
21779.27
Load Diagram
Shear Diagram
Moment Diagram
Shearing force at base = 960.85 kN
Bending moment at base = 21779.27 kN.m
3- Check overturning
Mbase U.L. = 21779.27 kN.m
Mbase U.L. 21779.27
Moverturning = 1.40
=
= 15556.62 kN.m
1.40
B
28768
2 =
+
+
Resisting Moment = WTotal
Resisting Moment
21 302064 kN.m
2 =
302064
Factor Of Safety = Over Turning Moment = 15556.62 = 19.4 > 1.5
Safe
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Page No.
Example.
The shown figure of a building which lies at Cairo. The building
consist of two basements, ground floor and mezzanine used as
hospital and 8 repeated floors used as residential building.Live load
and floor heights are shown in elevation, the average slab thickness
2
is 215 mm and Floor cover + walls = 2.5 kN/m . The soil is weak.
Due to Earthquake loads , it is required to :
1- Calculate the equivalent seismic load acting on the building .
2- Draw lateral load ,shear and over turning moment diagram over
the height of the structure.
2
L.L = 2.0 kN/m
2
L.L = 4.0 kN/m
2
2
L.L = 4.0 kN/m
2
L.L = 4.0 kN/m
2
6.0
L.L = 4.0 kN/m
2
L.L = 4.0 kN/m
6.0
3.0 +8 = 24 m
L.L = 4.0 kN/m
2
L.L = 4.0 kN/m
2
Void
6.0
2
L.L = 8.0 kN/m
6.0
2
L.L = 8.0 kN/m
2
L.L =10.0 kN/m
6.0
2
L.L =10.0 kN/m
6.0
6.0
6.0
6.0
6.0
Plan
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Sec . Elev
)
Page No.
3.5 3.5 4.0 4.0
L.L = 4.0 kN/m
Solution
1- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
Fig. (8-1)
P. (1/5 code)
Cairo
Table (8-1)
P. (2/5 code)
Weak soil
Table (8-2)
P. (2/5 code)
Zone (3)
a g= 0.15 g
Soil type (D)
Soil type (D)
Table (8-3)
P. (3/5 code)
Response spectrum curve Type (1)
T1 = Ct H
Cores
S =
TB =
TC =
TD =
3/4
C t = 0.05
P. (5/5 code)
Total Height of building (H) = 39 m
3/4
+
T1 = 0.05 39
Check
= 0.78sec.
T1 < ( 4 T C = 1.2 sec.)
O.K.
T1 < 2.0 sec.
O.K.
TC < T1 < TD
P. (3/5 code)
Sd (T1) = a g g1 S 2.5
R
R = 5.00
h = 1.00
Eq. (8-13)
TC h
> 0.20 a g g1
T1
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Page No.
1.80
0.10
0.30
1.20
Table (8-9)
P. (4/5 code)
Table (8-9)
P. (4/5 code)
g1 = 1.40
g1 = 1.40
g1
1.80
+
+
Sd (T1) = 0.15 g 1.40
+
More critical
0.20 a g g1 = 0.20
0.30
0.78
2.5
5.00
+
NOTE
g1 = 1.00
1.00 = 0.0727 g
+
+
0.15 g 1.40 = 0.042 g < Sd (T1)
O.K.
T1 > 2 TC = 0.60 sec
l = 1.00
P. (5/5 code)
+
ws = D.L. + a L.L.
+
ws = ( t s gc + F.C. + Walls ) + a L.L.
Table (8-7)
P. (5/5 code)
Table (8-7)
P. (5/5 code)
a = 0.25
a = 0.50
+
+
wfloor = ( D.L. + a L.L. ) area
+
D.L. = t s gc + F.C. + Walls = ( 0.215 25 + 2.5) = 7.875 kN/m2
+
+
Area = 30 30 - 6.0 6.0 = 864 m2
void
+
+
+
wfloor = ( 7.875 + 0.25 2.0 ) 864 = 7236 kN
+
Floor 12
+
+
+
wfloor = ( 7.875 + 0.50 10.0 ) 864 = 11124 kN
Floor 2 & 3 wfloor = ( 7.875 + 0.50 8.0 ) 864 = 10260 kN
Floor 4
11 wfloor = ( 7.875 + 0.25 4.0 ) 864 = 7668 kN
+
Floor 1
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+
+
+
+
WTotal = ( 11124 1 + 10260 2 + 7668 8 + 6912 1 )
= 100224 kN
Page No.
NOTE
L.L.
~
l W
g
+
= 0.0727 g
1.0
+
+
Fb = Sd (T1)
a
100224
g
F b = 7286.28 kN
2- Distribution of lateral force on each floor
+
Fi = Fb
w i Hi
w i Hi
i=n
i=1
Floor
No. Hi (m)
12
39.0
wi (kN)
wi H i
Fi (kN)
Q i (kN)
Mi (kN.m)
7236
282204
993.30
993.30
2979.90
11
36.0
7668
276048
971.63
1964.93
8874.68
10
33.0
7668
253044
890.66
2855.59
17441.46
9
30.0
7668
230040
809.69
3665.28
28437.30
8
27.0
7668
207036
728.72
4394.01
41619.32
7
24.0
7668
184032
647.75
5041.76
56744.60
6
21.0
7668
161028
566.78
5608.54
73570.24
5
18.0
7668
138024
485.82
6094.36
91853.31
4
15.0
7668
115020
404.85
6499.21
117850.14
3
11.0
10260
112860
397.24
6896.45
145435.94
2
7.0
10260
71820
252.79
7149.24
170458.28
1
3.5
11124
38934
137.04
7286.28
195960.26
i=n
i=1
w i Hi
i=n
i=1
Fi
= 2070090 = 7286.28
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Page No.
993.30
1964.93
2855.59
12
993.30
971.63
890.66
809.69
728.72
11
10
9
3665.28
4394.01
5041.76
8
647.75
7
6
5608.54
6094.36
566.78
485.82
5
6499.21
404.85
4
6896.45
397.24
3
7149.24
252.79
2
7286.28
137.04
1
Load Diagram
Shear Diagram
2979.90
8874.68
17441.46
28437.30
41619.32
56744.60
73570.24
91853.31
117850.14
145435.94
170458.28
12
11
10
9
8
7
6
5
4
3
2
1
195960.26
Moment Diagram
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Page No.
Example.
The shown figure of a student house which lies at Alexandria. The
building consist of 15 repeated floors & Floor height = 3 m. The
building consist of flat slab with average thickness 200 mm and
shear walls 300 mm thickness .The soil is medium dense sand.
It is required to :
1- Calculate the static wind load acting on the building in
Y - direction and show its distribution over the height .
2- Calculate the equivalent seismic load acting in Y- direction.
3- Draw lateral load ,shear and over turning moment diagram over
the height of the structure due to the critical lateral load .
4- Design the cross-section at the base of the Shear wall on axis 3
and draw its details of reinforcement .
Given that :
L.L. = 4.0 kN/m 2
1
2
3
6.0
4
4.0
6.0
12.0
Walls = 1.5 kN/m2
F.C. = 1.5 kN/m 2
6.0
5
4.0
6
6.0
6.0
6.0
Y
6.0
Plan
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Page No.
Solution
1- The static wind load
Ce = 0.8 + 0.5 = 1.3
V = 36 m/sec
Alexandria
2
+
2
+ + +
1.00 0.81 = 1.053 kN/m
1.15 0.81 = 1.211 kN/m2
1.40 0.81 = 1.474 kN/m 2
1.60 0.81 = 1.685 kN/m2
+
q = 1.3
q = 1.3
q = 1.3
q = 1.3
+ + +
+
+
+
k1
k2
k3
k4
+
+
+
+
+
F4 = P4 ( h b)
=1.685 15 30 = 758.25 kN
758.25 kN
1.685
kN/m
+
+
+
+
F3 = P3 ( h b)
=1.474 10 30 = 442.20 kN
10 m
+
kN/m
37.5
+
F1 = P1 ( h b)
=1.053 10 30 = 315.90 kN
1.211
363.30 kN
kN/m2
+
+
1.4742
442.20 kN
+
+
+
F2 = P2 ( h b)
=1.211 10 30 = 363.30 kN
+
15 m
2
10 m
25
Total wind force = 315.90 + 363.30
15
315.90 kN
+ 442.20 + 758.25
1.053 10 m
2
5
kN/m
= 751.86 kN
+
Fi H i
+
+
+
+
Total Moment at base ( Overturning Moment ) =
= F1 H1 + F2 H2 + F3 H3 + F4 H4
+
+
+
+
= 315.90 5 + 363.30 15 + 442.20 25 + 758.25 37.5
= 46518.375 kN.m
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Page No.
45 m
+
+
+
+
P1 = Ce
P2 = Ce
P3 = Ce
P4 = Ce
+
+
+
+
-4
0.5 r V C t Cs
2
2
q=
= 6.25 10 36 = 0.81 kN/m
1000
Zone A ( More Critical )
Pe = Ce k q
(kN/m2 )
2- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
Fig. (8-1)
P. (1/5 code)
Alexandria
Table (8-1)
P. (2/5 code)
Medium dense sand
Table (8-2)
P. (2/5 code)
Zone (2)
Soil type (C)
Soil type (C)
Table (8-3)
P. (3/5 code)
Response spectrum curve Type (2)
T1 = Ct H
a g= 0.125 g
S =
TB =
TC =
TD =
3/4
Shear walls
P. (5/5 code)
C t = 0.05
Total Height of building (H) = 45 m
+
T1 = 0.05
Check
3/4
45
= 0.87sec.
T1 < ( 4 T C = 2.4 sec.) O.K.
T1 < 2.0 sec.
TC < T1 < TD
O.K.
P. (3/5 code)
Sd (T1) = a g g1 S 2.5
R
R = 5.00
h = 1.00
Table (8-9)
P. (4/5 code)
Eq. (8-13)
TC h
> 0.20 a g g1
T1
g1 = 1.00
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Page No.
1.25
0.20
0.60
2.00
0.60
0.87
2.5
5.00
+
1.25
+
+
+
Sd (T1) = 0.125 g 1.00
1.00
= 0.0539 g
0.20 a g g1 = 0.20
+
+
0.125 g 1.00 = 0.025 g < Sd (T1)
O.K.
T1 < 2 TC = 1.20 sec
l = 0.85
P. (5/5 code)
+
+
ws = D.L. + a L.L.
ws = ( t s gc + F.C. + Walls ) + a L.L.
Table (8-7)
P. (5/5 code)
a = 0.25
+
+
4.0 = 9.0 kN/m2
+
+
+
ws = ( 0.20 25 + 1.5 + 1.5 ) + 0.25
wFloor = 9.0 12 30 = 3240 kN
wTotal = 3240 15 = 48600 kN
l W
g
+
= 0.0539 g
0.85
+
+
Fb = Sd (T1)
48600
g
F b = 2226.61 kN
3- Distribution of lateral force on each floor
+
Fi = Fb
w i Hi
w i Hi
i=1
i=n
w i Hi
H
= i=n i
w i Hi i = 1 Hi
i=1
i=n
‫ﻷن وزن اﻟ ﺪور ﺛﺎﺑ ﺖ‬
i = 15
i=1
H = 45 + 42 + 39 + 36 +33 + 30 + 27+ 24 + 21+ 18 + 15+ 12 + 9 + 6
i
+ 3 = 360 m
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Page No.
Fi
Fi
1 H
)
(
2226.61
=
360 i
=
Floor H (m) F (kN)
i
i
No.
15 45.0 278.33
6.185 H i
Moverturning =
Mbase U.L.
1.40
Q i (kN)
Mi (kN.m)
278.33
834.98
14
42.0
259.77
538.10
2449.27
13
39.0
241.22
779.31
4787.21
12
36.0
222.66
1001.97
7793.14
11
33.0
204.11
1206.08
11411.38
10
30.0
185.55
1391.63
15586.27
9
27.0
167.00
1558.63
20262.15
8
24.0
148.44
1707.07
25383.35
7
21.0
129.89
1836.95
30894.21
6
18.0
111.33
1948.28
36739.07
5
15.0
92.78
2041.06
42862.24
4
12.0
74.22
2115.28
49208.08
3
9.0
55.67
2170.94
55720.92
2
6.0
37.11
2208.05
62345.08
1
3.0
18.56
2226.61
69024.91
= 69024.91
= 49303.51 kN.m
1.40
Overturning Moment ( Seismic )
Overturning Moment ( wind )
= 49303.51 kN.m
= 46518.375 kN.m
The case of seismic load is the critical one
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Page No.
278.33
538.10
779.31
1001.97
1206.08
1391.63
1558.63
1707.07
1836.95
1948.28
2041.06
2115.28
2170.94
2208.05
2226.61
278.33
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
259.77
241.22
222.66
204.11
185.55
167.00
148.44
129.89
111.33
92.78
74.22
55.67
37.11
18.56
Load Diagram
Shear Diagram
15
834.98
14
2449.27
13
4787.21
12
7793.14
11
11411.38
10
15586.27
9
20262.15
8
25383.35
7
30894.21
6
36739.07
5
42862.24
4
49208.08
3
55720.92
2
62345.08
1
69024.91
Moment Diagram
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Page No.
4- Design of shear wall
+
Isw2 =
3
0.30 ( 6.0 )
12
1
2
4
= 5.40 m = Isw5
3
sw2
4
sw3
5
sw4
6
sw5
C.R.
3
+
Isw3 = 0.30 ( 4.0 ) = 1.60 m4 = Isw4
12
i=4
4
I = 2 ( 5.4 + 1.6 ) = 14.0 m
C.M.
e min
i=1
6.0
For Shear Walls on Axis 3
6.0
6.0
6.0
6.0
+
e min = 0.05 L = 0.05 30 = 1.5 m
+
+
1.6 3.0
2
2
2 ( 1.6 3.0 + 5.4 9.0 )
1 1.5
+
+
+
+
1+
emin)
+
+
1.6
= 2 ( 1.6 + 5.4 )
+
% of Sw3
Ii xi
(1
2
I ( xi)
i=1 i
i=n
+
+
% of each Shear Wall = i = In i 1 +
Ii
i=1
= 0.1223 = 12.23 %
Msw = % of Sw3 M base
3
+
3
+
= 0.01223 69024.91 = 8438.56 kN.m
+
ND.L. = g s
Area + o.w
+
By Area Method get the Normal force
8.0 4.0
No of floors
g s = ( t s gc + F.C. + Walls )
2
(
0.20
25
+
1.5
+
1.5
)
8.0
kN/m
=
=
+
+
+
+
15 = 7110 kN
No of floors
(6.0 8.0 )
+
+
= 4.0
Area
+
+
NL.L = Ps
No of floors
(6.0 8.0 ) + (0.3 4.0 3.0 25 )
+
+
ND.L. = 8.0
Area + o.w
+
+
ND.L. = g s
+
+
6.0
15 = 2880 kN
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Page No.
Nult = 1.12 ND.L.+ a N L.L.+ N S.L.
7110 + 0.25 2880 + 0 = 8683.2 kN
+
+
= 1.12
M ult = 1.12 M D.L.+ a M L.L.+ M S.L.
= 0 + 8438.56 = 8438.56 kN.m
+
+
+ + + +
Nu
= 8683.2 10 3 = 0.289
25 300 4000
Fcu b t
6
Mu
8438.56 10
0.0703
2 =
2 =
25 300 4000
Fcu b t
Get
-4
Fcu 10 ) b t
+
+
+
+
-4
= 2.0 25 10
300 4000 = 6000 mm2
+
\
As = A s = (
= 2.0
+
x = 0.9
+
As = 0.6 Ac= 0.6 ( 300 4000 ) = 7200mm2
min 100
100
12
5
8/m
5
10 / m \
5
16 / m \
28
12
12 28
28
0.3
4.0
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Page No.
Example.
For the given plan of a residential building, located in Hurghada
(Seismic Zone 5A). The building consists of ground floor and
5 typical floors. The building is constructed on a weak soil.
It is required to :
1- Calculate the shear base at ground floor level
2- Draw the lateral load & shear distribution diagrams
3- Calculate the seismic loads acting on the frame (F1)
Given that :
+
St. = 360/520
Fcu = 25 N/mm 2
F.C. + Partitions = 3.0 kN/m 2
t s av = 250 mm
All columns are (300 800)
The stiffness of exterior frames is twice the stiffness of interior
frames
7.0
4.0 4.0 4.0 4.0
7.0
F2
F1
L.L. = 2.0 kN/m 2
L.L. = 2.0 kN/m 2
L.L. = 2.0 kN/m 2
L.L. = 7.0 kN/m 2
L.L. = 7.0 kN/m 2
8.0
F3
8.0
F4
6.0
8.0
4.0
7.0
L.L. = 2.0 kN/m 2
8.0
Sec . Elev
Plan
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8.0
)
Page No.
Solution
1- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
Hurghada (Zone 5A)
Weak soil
Table (8-2)
P. (2/5 code)
Table (8-1)
P. (2/5 code)
a g = 0.25 g
Soil type (D)
Soil type (D)
Table (8-3)
P. (3/5 code)
Response spectrum curve Type (1)
T1 = Ct H
S =
TB =
TC =
TD =
3/4
Beams + Columns
R.C. Frames
P. (5/5 code)
C t = 0.075
Total Height of building (H) = 30 m
3/4
+
T1 = 0.075 30
Check
1.80
0.10
0.30
1.20
= 0.96 sec.
T1 < ( 4 T C = 1.2 sec.)
O.K.
T1 < 2.0 sec.
O.K.
TC < T1 < TD
P. (3/5 code)
Sd (T1) = a g g1 S 2.5
R
R = 5.00
h = 1.00
Eq. (8-13)
TC h
> 0.20 a g g1
T1
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Page No.
Table (8-9)
P. (4/5 code)
2.5
5.00
0.30
0.96
+
1.80
+
+
+
Sd (T1) = 0.25 g 1.00
g1 = 1.00
1.00
= 0.0703 g
0.20 a g g1 = 0.20
+
+
0.25 g 1.00 = 0.050 g < Sd (T1)
O.K.
T1 > 2 TC = 0.60 sec
l = 1.00
P. (5/5 code)
+
+
ws = D.L. + a L.L.
ws = ( t s gc + F.C. + Walls ) + a L.L.
Table (8-7)
P. (5/5 code)
a = 0.25
+
+
wfloor = ( D.L. + a L.L. ) area
+
D.L. = t s gc + F.C. + Walls = ( 0.25 25 + 3.0) = 9.25 kN/m2
+
+
Area = 16 21 - 4.0 7.0 = 308 m2
void
6
wfloor = ( 9.25 + 0.25 7.0 ) 308 = 3388 kN
wfloor = ( 9.25 + 0.25 2.0 ) 308 = 3003 kN
+
3
2
+
Floor
&
+
1
+
Floor
+
+
WTotal = ( 3388 2 + 10260 4 ) = 18788 kN
l W
g
+
= 0.0703 g
1.0
+
+
Fb = Sd (T1)
18788
g
F b = 1320.80 kN
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Page No.
2- Distribution of lateral force on each floor
+
Fi = Fb
Floor
No. Hi (m)
6
30.0
w i Hi
w i Hi
i=1
i=n
wi (kN)
wi H i
Fi (kN)
Q i (kN)
3003
90090
327.96
327.96
5
26.0
3003
78078
284.23
612.19
4
22.0
3003
66066
240.50
852.69
3
18.0
3003
54054
196.77
1049.46
2
14.0
3388
47432
172.67
1222.13
1
8.0
3388
27104
98.67
1320.80
i=6
i=1
w i Hi
= 362824
327.96
612.19
852.69
1049.46
1222.13
1320.80
6
5
4
3
2
1
Shear Diagram
327.96
284.23
240.50
196.77
172.67
98.67
Load Diagram
3- Seismic loads acting on frame (F1)
Plan is unsymmetric as C.M. = C.R.
- C.R. is at the center of the plan as K is symmetric for frames
( K F1 = K F4 & K F2 = K F3 )
21
X C.R.= 2 = 10.5 m
- As the stiffness of exterior frames is twice the stiffness of interior
frames
( K F1 = K F4 = 2 K F2 = 2 K F3 )
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Page No.
7.0
14.0
4.0
Datum
A2
C.G.
12.0
A1
C.M.
C.G.
X C.M.
3.5
14.0
Ai x i
Ai
+
+
+
+
wi Ai x i =
wi A i
X C.M.=
+
+ +
+
+
X C.M.= 7.0 12.0 3.5 + 14.0 16.0 14.0 = 11.14 m
+
7.0 12.0 + 14.0 16.0
NOTE
D.L. & L.L.
wi
( 1 e* )
+
KFi xi
i=n
2
K
(
x
)
i
Fi
i=1
+
1 +
+
Fi
% of each Frame = i = K
n
K Fi
i=1
+
+
e = X C.M. - X C.R. = 11.14 - 10.5 = 0.64 m
e * = e + 0.05 L = 0.64 + 0.05 21.0 = 1.69 m
K F1 = KF4 & KF2 = K F3
K F1 = 2 K F2
i=1
K Fi = 2 K F1 + 2 K F2 = 2 2 K F2 + 2 K F2 = 6 K F2
+
i=n
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Page No.
7.0
8.0
7.0
7.0
MT
0.64
C.R. C.M.
Zone (2)
Zone (1)
8.0
F1
3.5
+
+
+
+
1 1.69
+
1 1.69
+
+
+
2 K F2 10.5
2
2
2 ( 2 K F2 10.5 + K F2 3.5 )
+
1+
2 ( K F1
K F1 10.5
2
2
10.5 + K F2 3.5 )
+
2 K F2
6 K F2
‫ ﯾﻘ ﻊ ﻓ ﻲ‬frame ‫ﻷن‬
+
=
1+
zone (1)
+
K F1
6 K F2
+
% of F1 =
+
10.5
= 0.4096 = 40.96 %
+
+
Fi = % F1 Fi = 0.4096 Fi
for F1
134.33
116.42
98.51
80.60
70.72
40.41
Load Diagram
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Page No.
Example.
The figure shows a residential building which lies at Aswan. It
consist of 15 repeated floors & Floor height = 3 m. The building
consist of prestressed slab with average thickness 200 mm and
shear walls and cores 250 mm thickness .The soil is weak.
It is required to :
1- Calculate the equivalent static load acting on the building at
each floor and show its distribution over the height .
2- Determine the center of rigidity of the structure .
3- Compute the torsion moment at the ground level.
4- Calculate the percentage of lateral load acting on each shear
wall and core.
Given that :
L.L. = 6.0 kN/m 2
1
Walls = 1.5 kN/m2
F.C. = 1.5 kN/m 2
2
3
4
5
6
A
5.0
B
5.0
C
5.0
D
5.0
E
6.0
6.0
6.0
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6.0
)
6.0
Page No.
Solution
1- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
Fig. (8-1)
P. (1/5 code)
Aswan
Weak soil
Table (8-2)
P. (2/5 code)
Zone (2)
Table (8-1)
P. (2/5 code)
a g = 0.125 g
Soil type (D)
Soil type (D)
Table (8-3)
P. (3/5 code)
Response spectrum curve Type (1)
T1 = Ct H
S =
TB =
TC =
TD =
3/4
Shear walls + Cores
C t = 0.05
P. (5/5 code)
Total Height of building (H) = 45 m
Check
3/4
+
T1 = 0.05
45
= 0.87sec.
T1 < ( 4 T C = 1.2 sec.)
O.K.
T1 < 2.0 sec.
O.K.
TC < T1 < TD
P. (3/5 code)
Sd (T1) = a g g1 S 2.5
R
R = 5.00
Prestressed concrete
Eq. (8-13)
TC h
> 0.20 a g g1
T1
Table (8-4)
P. (3/5 code)
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h = 1.05
)
Page No.
1.80
0.10
0.30
1.20
Table (8-9)
P. (4/5 code)
0.30
0.87
2.5
5.00
+
1.80
+
+
+
Sd (T1) = 0.125 g 1.00
g1 = 1.00
1.05
= 0.0407 g
0.20 a g g1 = 0.20
+
+
0.125 g 1.00 = 0.025 g < Sd (T1)
O.K.
T1 > 2 TC = 0.60 sec
l = 1.00
P. (5/5 code)
+
+
ws = D.L. + a L.L.
ws = ( t s gc + F.C. + Walls ) + a L.L.
Table (8-7)
P. (5/5 code)
a = 0.25
+
+
ws = ( 0.20 25 + 1.5 + 1.5 ) + 0.25
6.0 = 9.5 kN/m2
+
+
Fb = Sd (T1)
l W
g
+
= 0.0407 g
1.0
+
+ +
wFloor = 9.5 20 30 = 5700 kN
wTotal = 5700 15 = 85500 kN
85500
g
F b = 3479.85 kN
2- Distribution of lateral force on each floor
+
Fi = Fb
w i Hi
w i Hi
i=1
i=n
w i Hi
H
= i=n i
w i Hi i = 1 Hi
i=1
i=n
‫ﻷن وزن اﻟ ﺪور ﺛﺎﺑ ﺖ‬
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Page No.
Floor H (m)
i
No.
i = 15
H = 45 + 42 + 39 + 36 +33 + 30 + 27
i
+ 24 + 21+ 18 + 15+ 12 + 9 + 6 + 3
= 360 m
i=1
Fi
Fi
Fi (kN)
15
45.0
434.98
14
42.0
405.98
13
39.0
376.98
=
1 H
( 3479.85) 360
i
12
36.0
347.99
11
33.0
318.99
=
9.666 H i
10
30.0
289.99
9
27.0
260.99
8
24.0
231.99
7
21.0
202.99
6
18.0
173.99
5
15.0
144.99
4
12.0
116.00
3
9.0
87.00
2
6.0
58.00
1
3.0
29.00
434.98
405.98
376.98
347.99
318.99
289.99
260.99
231.99
202.99
173.99
144.99
116.00
87.00
58.00
29.00
Load Diagram
3- Center of rigidity and torsional moment
1
For Shear walls
3
4
5
6
a
3
5.0
+
Ix = 0.25 10.0
2
b
12
5.0
C.R.
4
= 20.83 m
c
x
10.0
C.M.
x
5.0
d
5.0
e
6.0
6.0
6.0
6.0
0.25
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Page No.
6.0
For Core
y
x
5.0
+
+ +
+
+ +
+
+
y = 4.75 0.25 2 2.625 + 6.0 0.25 0.125
4.75 0.25 2 + 6.0 0.25
0.25
= 1.66 m
6.0
( 2.625 - 1.66 )
3
2
2
4
+
+
+
20.83 0 + 20.83 6 + 10.22 21 + 20.83 30
20.83 3 + 10.22
+ +
+
+
+ 0.25 6.0 ( 1.66 - 0.125 ) = 10.22 m
Ii xi =
Ii
X C.R.=
+
+
+ 0.25 4.75
+
+ 6.0 0.25
12
2
+
+
3
+
0.25 4.75
12
Ix =
x
= 13.26 m
X C.M.= L = 30 =15.0 m
2
2
+
e = X C.M. - X C.R. = 15.0 - 13.26 = 1.74 m
e * = e + 0.05 L = 1.74 + 0.05 30.0 = 3.24 m
4- Percentage of lateral load carried by shear walls & cores
1
2
3
4
5
+
+
i=n
+
Ii xi
(1
2
I ( x i)
i=1 i
% of each Shear Wall = i = nI i 1 +
Ii
i=1
e* )
6
a
5.0
b
5.0
C.R.
c
C.M.
5.0
d
(+ ve)
Zone (1)
(- ve)
Zone (2)
5.0
e
6.0
6.0
6.0
6.0
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6.0
)
Page No.
+
+
20.83 13.26
2
2
2
2
20.83 (13.26 + 7.26 + 16.74 ) + 10.22 7.74
+
+
1 - ( 1 3.24 )
+
F 1 =20.8320.83
3 + 10.22
+
For Shear Walls on Axis 1
= 20.66 %
+
20.83 7.26
2
2
2
20.83 (13.26 + 7.26 + 16.74 ) + 10.22 7.74
+
2
+
+
1 - ( 1 3.24 )
+
F 2 =20.8320.83
3 + 10.22
+
For Shear Walls on Axis 2
= 24.28 %
+
20.83 16.74
2
2
2
20.83 (13.26 + 7.26 + 16.74 ) + 10.22 7.74
1 + (1 3.24 )
10.22 7.74
2
2
2
20.83 (13.26 + 7.26 + 16.74 ) + 10.22 7.74
+
+
1 + ( 1 3.24 )
+
2
+
+
F 3 =20.8320.83
3 + 10.22
+
For Shear Walls on Axis 6
= 38.73 %
+
+
2
+
+
F4 =20.8310.22
3 + 10.22
+
For Core
= 16.34 %
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Page No.
Example.
The shown figure of a residential building which lies at Cairo . It
consist of a ground, mezzanine and 8 repeated floors and two
basements. Live load and floor heights are shown in elevation.
The soil below the building is weak.
Due to Earthquake loads , it is required to :
1- Calculate the story shear at each floor level .
2- Calculate the total lateral drift .
3- Draw the distribution of web drift along the height of the building .
Given that :
F.C. = 1.5 kN/m 2
Slab thickness = 160 mm
All Beams are ( 300 600 )
All Column are ( 300 700 )
+
+
2
E = 22100 N/mm
L.L = 2.0 kN/m 2
L.L = 4.0 kN/m 2
L.L = 4.0 kN/m 2
+
5
6 m =30m
L.L = 4.0 kN/m 2
4+ 6m =24 m
L.L = 4.0 kN/m 2
L.L = 4.0 kN/m 2
L.L = 4.0 kN/m 2
L.L = 4.0 kN/m 2
L.L = 8.0 kN/m
2
L.L = 8.0 kN/m 2
Plan
L.L = 10.0 kN/m 2
L.L = 10.0 kN/m 2
Sec . Elev
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Page No.
3.0 +12 = 36 m
L.L = 4.0 kN/m 2
Solution
1- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
Fig. (8-1)
P. (1/5 code)
Aswan
Table (8-1)
P. (2/5 code)
Loose sand
Table (8-2)
P. (2/5 code)
Zone (2)
a g = 0.125 g
Soil type (D)
Soil type (D)
Table (8-3)
P. (3/5 code)
Response spectrum curve Type (1)
T1 = Ct H
S =
TB =
TC =
TD =
3/4
Beams + Columns
R.C. Frames
P. (5/5 code)
C t = 0.075
Total Height of building (H) = 36 m
3/4
+
T1 = 0.075 36
Check
1.80
0.10
0.30
1.20
= 1.10sec.
T1 < ( 4 T C = 1.2 sec.)
O.K.
T1 < 2.0 sec.
O.K.
TC < T1 < TD
P. (3/5 code)
Sd (T1) = a g g1 S 2.5
R
Eq. (8-13)
TC h
> 0.20 a g g1
T1
R = 5.00
h = 1.00
Table (8-9)
P. (4/5 code)
g1 = 1.00
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Page No.
2.5
5.00
0.30
1.10
+
1.80
+
+
+
Sd (T1) = 0.125 g 1.00
1.00
= 0.0307 g
0.20 a g g1 = 0.20
+
+
0.125 g 1.00 = 0.025 g < Sd (T1)
O.K.
T1 > 2 TC = 0.60 sec
l = 1.00
P. (5/5 code)
+
+
ws = D.L. + a L.L.
ws = ( t s gc + F.C. + Walls ) + a L.L.
Table (8-7)
P. (5/5 code)
a = 0.25
+
+
+
+
+
+
+
+
+
+
4
+
3.0 = 6520.50 kN
+
24 30 + 0.3 0.6 25 ( 24 6 )
wFloor = ( 5.5 + 0.25 8.0 )
+ 0.3 0.7 25 30
Floor
+
+
+
+
C.L.-C.L.
3
+
&
+
2
+
Floor
+
+
+
+
ws = ( 0.16 25 + 1.5 ) = 5.5 kN/m2
Floor 1
beams
slab
wFloor = ( 5.5 + 0.25 10.0 ) 24 30 + 0.3 0.6 25 ( 24 6)
columns
+ 0.3 0.7 25 30 4.5 = 7116.75 kN NOTE
11
+
+
+
+
+
+
+
+
+
+
+
wFloor = ( 5.5 + 0.25 4.0 ) 24 30 + 0.3 0.6 25 ( 24 6)
+ 0.3 0.7 25 30 3.0 = 5800.5 kN
+
)
+
+
+
+
+
+
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+
1.5 = 5204.25 kN
wFloor = ( 5.5 + 0.25 2.0 )
+ 0.3 0.7 25 30
+
+
24 30 + 0.3 0.6 25 ( 24 6)
+
Floor 12
Page No.
+
+
l W
g
+
= 0.307 g
1.0
+
+
Fb = Sd (T1)
+
+
WTotal = ( 7116.75 1 + 6520.50 2 + 5800.50 8 + 5204.25 1 )
= 71766 kN
71766
g
F b = 2203.22 kN
2- Distribution of lateral force on each floor
+
Fi = Fb
Floor
No. Hi (m) wi (kN)
36.0 5204.25
12
w i Hi
w i Hi
i=1
i=n
wi H i
Fi (kN)
Fi H i (kN.m)
187353.00
305.63
11002.57
11
33.0
5800.50
191416.50
312.26
10304.44
10
30.0
5800.50
174015.00
283.87
8516.06
9
27.0
5800.50
156613.50
255.48
6898.01
8
24.0
5800.50
139212.00
227.09
5450.28
7
21.0
5800.50
121810.50
198.71
4172.87
6
18.0
5800.50
104409.00
170.32
3065.78
5
15.0
5800.50
87007.50
141.93
2129.02
4
12.0
5800.50
69606.00
113.55
1362.57
3
9.0
6520.50
58684.50
95.73
861.58
2
6.0
6520.50
39123.00
63.82
382.93
1
3.0
7116.75
21350.25
34.83
104.49
i = 12
i = 12
w i Hi
i=1
i=1
= 1350600.75
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Fi H i
= 54250.59
)
Page No.
3- Total drift
Chord Drift
Fi H i = 54250.59 kN.m
M over turning =
2
= 54250.59 = w 36
2
w = 83.72 kN/m
+
M =wH
2
2
3
+
Ic = 0.3 0.7 = 0.0086 m4
12
12.0
( I c + A 12 2 )
+
+
( Ic + A 6 2 ) + 2
+
[ Ic + 2
+
Ie = 6
6.0
]
+
+
+
+
+
+
2
2
= 6 [0.0086 + 2 (0.0086 + 0.3 0.7 6 ) + 2 (0.0086 + 0.3 0.7 12 )]
4
= 453.86 m
4
4
= 0.00175 m
+
+
+
+
wH =
83.72 36
Chord drift (Dc) = 8EI
8 2.21 10 7 453.86
e
= 1.75 mm
Web Drift
Assume ( K ) constant for all floors
Typical Story
3
+
4
Ic = 0.3 0.7 = 0.0086 m
12
3
+
4
Ib1= Ib2 = 0.3 0.6 = 0.0054 m
12
12EI c1
h3
7
1
2 0.0086
1+
3 (0.0054 +0.0054 )
6.0
6.0
= 20183.4 kN/m
+
( 3.0 )3
+
+
+
K int.=
12 2.21 10 0.0086
+
K int.=
1
2I c1
1+
h ( Ibb1 + Ibb2 )
1
2
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Page No.
1
2I c2
h ( Ibb1 )
1
12EI c2
1+
h3
K ext.=
7
+
+
+
3
+
( 3.0 )
= 11460.9 kN/m
+
=
1
2 0.0086
1+
3 (0.0054 )
6.0
12 2.21 10 0.0086
K = SK
+int.
SK
F1
ext. = 3 K int. + 2 K ext.
i=6
i=1
K Fi = 6 K F1 = 6 83472 = 500832 kN/m
+
Ks =
+
+
= 3 20183.4 + 2 11460.9 = 83472 kN/m
Floor
No. Hi (m)
12
36.0
S Qi
Fi (kN)
Q i (kN)
Qi
305.63
305.63
18083.53
36.11
Di =
Ks
(mm)
36.11
35.50
11
33.0
312.26
617.88
17777.90
35.50
10
30.0
283.87
901.75
17160.02
34.26
9
27.0
255.48
1157.23
16258.27
32.46
8
24.0
227.09
1384.33
15101.03
30.15
7
21.0
198.71
1583.04
13716.71
27.39
6
18.0
170.32
1753.36
12133.67
24.23
20.73
5
15.0
141.93
1895.29
10380.31
20.73
16.94
4
12.0
113.55
2008.84
8485.02
16.94
3
9.0
95.73
2104.57
6476.18
12.93
2
6.0
63.82
2168.39
4371.61
8.73
1
3.0
34.83
2203.22
2203.22
4.40
i = 12
i=1
Qi
= 18083.53
Total web drift ( SD i) =
S Qi
Ks
34.26
32.46
30.15
27.39
24.23
12.93
8.73
4.40
Web Drift
Diagram
18083.53
= 500832 = 0.03611 m = 36.11 mm
Total Drift = Web Drift + Chord Drift
= 36.11 + 1.75 = 37.86 mm.
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Page No.
Example.
The given figures show sectional elevation and cross sections
of a minaret of 51.4 meters height constructed in Sharm El Sheikh
above loose soil. The thickness of minaret walls is 400 mm up to
level 33.4 m supporting 8 circular posts of 300 mm diameter.
The rest of the minaret is a cylindrical wall of 300 mm thickness
covered by a cone of 3.0 height. The minaret is supported on
shallow foundation,where top of footing is at 2.0 meter depth
below ground.
It is required to :
1- Calculate the equivalent base shear acting on the minaret.
2- Determine the overturning moment of the minaret.
3- Calculate the breadth of the minaret footing to ensure proper
safety against over turning.
Given that :
- No. L.L. or F.C. will be considered in calculations
- A steel stair is used inside the minaret, its weight will be neglected.
- Foundation depth is 1.0 m. (Take foundation weight into
consideration in stability calculations).
- Surface area of the cone is as follows
Surface Area =
L r
+
L
+
h
C.G.
r
h/3
Cone
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Page No.
0.40
3.00
3.00
0.40
4.00
8.00
4.00
SEC. (A-A)
E
E
D
D 4.00
C
C
B
B 10.00
A
A
0.
40
1.66
8.00
4.00
0.40
0.40
2.60
0.3
0
1.66
4.00
SEC. (B-B)
6.00
2.49
1.76
SEC. (E-E)
1.76
1.66
1.76
0.
40
40
0.
1.90
Circular Posts
300mm
1.08
1.08
2.49
1.90
0.91
1.08
0.30
1.08
15.00
0.91
1.76
SEC. (D-D)
SEC. (C-C)
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Page No.
Solution
1- Simplified Modal Response Spectrum
+
Fb = Sd (T1)
l W
g
According to Soil type and building location
(8-1)
Sharm El Sheikh P.Fig.
(1/5 code)
Table (8-1)
P. (2/5 code)
Loose soil
(8-2)
Zone (5B) P.Table
(2/5 code)
Soil type (D)
S =
TB =
TC =
TD =
Soil type (D)
Table (8-3)
P. (3/5 code)
Response spectrum curve Type (1)
T1 = Ct H
a g = 0.30 g
1.80
0.10
0.30
1.20
3/4
R.C. Core
C t = 0.05
P. (5/5 code)
Total height of minaret from foundation (H) = 50.4 m
Check
3/4
+
T1 = 0.05
50.4
= 0.946 sec.
T1 < ( 4 T C = 1.2 sec.)
O.K.
T1 < 2.0 sec.
O.K.
TC < T1 < TD
P. (3/5 code)
Sd (T1) = a g g1 S 2.5
R
Table (A)
P. (4/5 code)
51.4
50.4
48.4
Eq. (8-13)
TC h
> 0.20 a g g1
T1
2.0
R = 3.50
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Page No.
Table (8-4)
P. (3/5 code)
Reinforced concrete
1.80
+
1.20
2.5
3.50
0.30
0.946
+
g1 = 1.20
+
Sd (T1) = 0.30 g
+
Table (8-9)
P. (4/5 code)
h = 1.00
1.00
= 0.1468 g
0.20 a g g1 = 0.20
+
+
0.30 g 1.20 = 0.072 g < Sd (T1)
O.K.
T1 > 2 TC = 0.60 sec
l = 1.00
P. (5/5 code)
Calculation of Minaret weight
Part (7)
Part (1)
2.94
+ +
Part (5)
7.84
3.92
Part (4)
+
+
+
+
+
+ +
+ + +
+
Volume
Density
Weight =
= ( Surface Area Thickness ) Density
= ( Perimeter h Thickness ) Density
= ( Perimeter h t w ) gc
= 4 4.0 17.0 0.40 25
= 2720 kN
Part (6)
4.0
7.84
0.39
50.4
Part (2)
0 .4
0
9.80
0.39
4.0
Part (3)
0.40
3.92
17.0
Part (1)
17.0
2.0
3.92
SEC. (A-A)
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Page No.
1.62
39
+
0.
+
+
+ +
+
+
Part (2) ( h = 10.0 m )
Weight = ( Perimeter h t w ) gc
= 8 1.66 10.0 0.40 25
= 1328 kN
3.92
0.40
1.62
3.92
SEC. (B-B)
1.72
2.44
1.72
+
+
Part (3) ( Horizontal slab t s = 0.40 m )
Weight = ( Area t s ) gc
2
+
+
+
+
+ 2.49 ) 1.9 0.40 25
( 0.912.0
1
8
= 258.4 kN
1.86
=8
2.44
3
1.86
4
0.89
0.89
7
6
5
SEC. (C-C)
+
1.06
+
+
+ +
+
39
+
1.06
0.
Part (4) ( h = 8.0 m )
Weight = ( Perimeter h t w ) gc
= 8 1.08 8.0 0.40 25
= 691.2 kN
SEC. (D-D)
0.29
+
+
+
Part (5) ( h = 4.0 m )
Weight = ( Area h No. of posts ) gc
2
0.3
4
+
+
+
+
4.0 8.0 25
=
= 226.2 kN
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Circular Posts
300mm
SEC. (D-D)
)
Page No.
+
2.60
0.3
0
+
+
+ +
+
+
Part (6) ( h = 8.0 m )
Weight = ( Perimeter h t w ) gc
2.60 8.0 0.30 25
=
= 490.1 kN
SEC. (E-E)
Part (7)
3.354
3.0
+
+
Surface Area =
(Given)
3.0
L
L r h
+
+
+
Volume
Density
Weight =
= ( Surface Area Thickness ) Density
r
Cone
+
+ +
+
+
+
+ + +
+
+
+
L r Thickness ) Density
=(
L r Thickness ) Density
=
3.354 1.50 0.30 25
=
= 118.5 kN
WTotal = WParts
WTotal = 2720 + 1328 + 258.4 + 691.2 + 226.2 + 490.1 + 118.5
WTotal = 5832.4 kN
+
= 0.1468 g
l W
g
1.0
+
+
Fb = Sd (T1)
5832.4
g
F b = 856.2 kN
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Page No.
2- Distribution of lateral force on each part
Part (7)
+
Fi = Fb
2.94
Part (6)
W6
7.84
Part (5)
W5
3.92
Part (4)
W7
w i Hi
w i Hi
i=n
i=1
Check Overturning
W4
Part (3)
7.84
48.4
W2
43.4
Overturning Moment
Part (2)
0.39
50.4
W3
9.80
37.4
31.4
27.2
Part (1)
22.0
W1
17.0
8.5
2.0
(Given) 1.0
Overturning Point
B
Part
No.
7
Hi (m)
wi (kN)
wi H i
48.4
118.5
5735.4
42.14
49.4
2081.68
6
43.4
490.1
21270.34
156.28
44.4
6938.74
5
37.4
226.2
8459.88
62.16
38.4
2386.82
4
31.4
691.2
21703.68
159.46
32.4
5166.56
3
27.2
258.4
7028.48
51.64
28.2
1456.24
2
22.0
1328
29216
214.66
23.0
4937.10
1
8.5
2720
23120
w i Hi
169.87
9.5
1613.74
i=7
i=1
Fi (kN) Hbase (m)
i=7
i=1
= 116533.78
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Fi H i (kN.m)
Fi Hbase
= 24580.89
)
Page No.
3- Check overturning
Mbase U.L. = 24580.89 kN.m
Mbase U.L. 24580.89
Moverturning = 1.40 = 1.40
= 17557.78 kN.m
+
Footing Dimensions = B B
0.40
+
+
+
Footing Weight = B B 1.0 25
2
= 25 B
2
WTotal = 25 B + 5832.4
0.40
4.00
4.00
+
Resisting Moment = WTotal
B
2
B
+
2
(
25
B
+ 5832.4 ) 2B
=
3
12.5
B
+ 2916.2 B
=
Resisting Moment
Factor Of Safety = Over Turning Moment
3
1.50 =
12.5 B + 2916.2 B
24580.89
B = 9.25 m
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Page No.
B
NOTE
Overturning Moment
Soil Stresses
Counter Weight
OR
Filling Material
(Plain Concrete)
as counter weight
Piles
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