( 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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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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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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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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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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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
)
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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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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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) C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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. C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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. C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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
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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
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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 ) C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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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
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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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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. C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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) C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
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 C Copyright . All copyrights reserved. Downloading or printing of these notes is allowed for personal use only. ( Commercial use of these notes is not allowed.
)
Page No.