response spectrum by egyptian code seismic load
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ASSIGNMENT NO. 2
Prepared By: Ahmed Shaban Mahmoud
Date : 24-Nov-2010
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Numerical Analysis Assignment NO.2 24-NOV-2010
Prepared By: Ahmed Shaban Mahmoud
Page 2 of 21
Q2. Draw the following curves in [Sae-T] and [Sae-Sde ] formats for both
response spectrum and Design Spectrum (Steel Frame with Welded
Connections)
Curve 1: All Soil Classes for ag=0.15g
Curve 2: All Soil Classes for ag=0.3g
Curve 3: All Design Ground Accelerations for Soil Class C
Curve 4: All Design Ground Accelerations for Soil Class B
Q3. Solve Question-of Exam-2009
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Numerical Analysis Assignment NO.2 24-NOV-2010
Prepared By: Ahmed Shaban Mahmoud
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Solution
Q2. Show below figures
Curve 1: All Soil Classes for ag=0.15g
Figure: Elastic Spectral Acc. VS. Period T for all type of soil
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
Elastic
SpectralAcc.
Sed
(g)
Period T (Sec.)
Acc. Spectrum (Elastic)
ag =0.15g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Figure: Design Spectral Acc. VS. Period T for all type of soil
0
0.1
0.2
0.3
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
Design
SpectralAcc.
Sad
(g)
Period T (Sec.)
Acc. Spectrum (Design)
ag =0.15g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Figure: Elastic Spectral Acc. VS. Elastic Spectral Disp. for all
type of soil
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
0.0000 0.0010 0.0020 0.0030 0.0040 0.0050 0.0060 0.0070 0.0080 0.0090 0.0100
ElasticSpectralA
cc.
Sae
(g)
Elastic Spectral Disp. Sde (m)
Acc. Spectrum (Elastic)
ag =0.15g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Figure: Design Spectral Acc. VS. Design Spectral Disp. for all
type of soil
Curve 2: All Soil Classes for ag=0.3g
0
0.05
0.1
0.15
0.2
0.25
0.0000 0.0020 0.0040 0.0060 0.0080 0.0100 0.0120 0.0140 0.0160
Design
Spec
tralAcc.
Sad
(g)
Design Spectral Disp. Sdd (m)
Acc. Spectrum (Design)
ag =0.15g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Numerical Analysis Assignment NO.2 24-NOV-2010
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Figure: Elastic Spectral Acc. VS. Period T for all type of soil
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
ElasticSpectralAcc.
Sed
(g)
Period T (Sec.)
Acc. Spectrum (Elastic)
ag =0.3g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Figure: Design Spectral Acc. VS. Period T for all type of soil
0
0.1
0.2
0.3
0.4
0.5
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
Design
Spectr
alAcc.
Sad
(g)
Period T (Sec.)
Acc. Spectrum (Design)ag =0.3g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Figure: Elastic Spectral Acc. VS. Elastic Spectral Disp. for all
type of soil
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
0.0000 0.0050 0.0100 0.0150 0.0200
ElasticSpectralAcc.
Sae
(g)
Elastic Spectral Disp. Sde (m)
Acc. Spectrum (Elastic)ag =0.3g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Figure: Design Spectral Acc. VS. Design Spectral Disp. for all
type of soil
Curve 3: All Design Ground Accelerations for Soil Class C
0
0.1
0.2
0.3
0.4
0.5
0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350
Design
Spec
tralAcc.
Sad
(g)
Design Spectral Disp. Sdd (m)
Acc. Spectrum (Design)
ag =0.3g , =1.2
Soil Class A
Soil Class B
Soil Class C
Soil Class D
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Figure: Elastic Spectral Acc. VS. Period T for all ground acc.
type of soil C
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
Elastic
SpectralAcc.
Sed
(g)
Period T (Sec.)
Acc. Spectrum (Elastic)
Soil Class C , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Figure: Design Spectral Acc. VS. Period T for all ground acc.type of soil C
0
0.1
0.2
0.3
0.4
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
Design
Sp
ectralAcc.
Sad
(g)
Period T (Sec.)
Acc. Spectrum (Design)
Soil Class C , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Numerical Analysis Assignment NO.2 24-NOV-2010
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Figure: Elastic Spectral Acc. VS. Elastic Spectral Disp. for all
ground acc. type of soil C
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0.0000 0.0020 0.0040 0.0060 0.0080 0.0100 0.0120 0.0140
ElasticSp
ectralAcc.
Sae
(g)
Elastic Spectral Disp. Sde (m)
Acc. Spectrum (Elastic)
Soil Class C , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Numerical Analysis Assignment NO.2 24-NOV-2010
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Figure: Design Spectral Acc. VS. Design Spectral Disp. for all
ground acc. type of soil C
Curve 4: All Design Ground Accelerations for Soil Class B
0
0.1
0.2
0.3
0.4
0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350
Design
Sp
ectralAcc.
Sad
(g)
Design Spectral Disp. Sdd (m)
Acc. Spectrum (Design)
Soil Class C , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Figure: Elastic Spectral Acc. VS. Period T for all ground acc.
type of soil B
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
Elastic
SpectralAcc.
Sed
(g)
Period T (Sec.)
Acc. Spectrum (Elastic)
Soil Class B , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Figure: Design Spectral Acc. VS. Period T for all ground acc.
type of soil B
0
0.1
0.2
0.3
0.4
0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2
Design
SpectralAcc.
Sad
(g)
Period T (Sec.)
Acc. Spectrum (Design)
Soil Class B , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Numerical Analysis Assignment NO.2 24-NOV-2010
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Figure: Elastic Spectral Acc. VS. Elastic Spectral Disp. for all
ground acc. type of soil B
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0.0000 0.0020 0.0040 0.0060 0.0080 0.0100 0.0120
ElasticSpectralAcc.
Sae
(g)
Elastic Spectral Disp. Sde (m)
Acc. Spectrum (Elastic)
Soil Class B , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Numerical Analysis Assignment NO.2 24-NOV-2010
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Figure: Design Spectral Acc. VS. Design Spectral Disp. for all
ground acc. type of soil B
Q3.1-Estimation Wind Force F4 according to ANSI
Basic wind speed (3-sec gust speed) = 100 mph = 44.704 m/s
0
0.1
0.2
0.3
0.4
0.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350
DesignS
pectralAcc.
Sad
(g)
Design Spectral Disp. Sdd (m)
Acc. Spectrum (Design)
Soil Class B , =1.2
Region 1
Region 2
Region 3
Region 4
Region 5.a
Region 5.b
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Important factor (Structure category) I = 1.07
Terrain exposure constant : for terrain exposure B
= 4.5 Zg = 1200 ft = 365.76 m Do = 0.01
Kz = 2.58 ( Z / Zg )1/
qz = 0.613 Kz ( V I )2 ( N/m2 )
= 0.613 Kz * ( 1.07 * 44.704 )2 ( N/m2 )
Guest Factor G is calculated accord. ANSI as shown below.
B = 15m building width
h = 75m building height
H / B =75 / 15 = 5.0 So,building is rigid
Tz =. ( )
.
(
.)
. ( . ).
(
.)
= 0.1472
G = 0.65 + 3.65 Tz
= 1.18736
L / B = 25 / 25 = 1.0
Cp = 0.8 for windward wall
Cp = -0.5 for leeward wall
Wind On building :-
Floor
NO.
Z
( m )KZ I
qz( N/m2 )
Cp
(front+rear)Gh
Pz( N/m2 )
5TH
75 1.81 1.07 2544.6 1.3 1.1874 3928
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Numerical Analysis Assignment NO.2 24-NOV-2010
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4th
60 1.73 1.07 2421.5 1.3 1.1874 3738
3rd
45 1.62 1.07 2271.5 1.3 1.1874 3506
AVG. 3724
Wind Force F4 = ( 3928+2*3738+3506 ) /4 * 25 * 15 /1000 = 1398 KN
Q3.2-Distributing Wind Force F4 = 1396 KN
By Upper Limit ( Rigid Diaphram )
Distribution is according to stiffness
Force in bracing at axis 1 or 3 = 8 / (8+1+8) * 1398 = 657.9 KN
Force in frame at axis 2 = 1 / (8+1+8) * 1398 = 82.2 KN
By Lower Limit ( NO-Diaphram )
Distribution is according to served area
Force in bracing at axis 1 or 3 = 1 / 4 * 1398 = 350 KN
Force in frame at axis 2 = 1 / 2 * 1398 = 699 KN
Q3.3-a. Seismic force according to Egypt code for loads
Check time period 0.8 sec 2.0 sec & 4 Tc (Tc atleast 0.25 ),and regular horizontal plan with regular vertical elevations.
So, Use simplified modal response spectrum method
Fb = Sd ( T1 ) . W / g
Where :-Sd ( T1 ) is design spectrum acceleration calculated for time period =0.8 sec- = 1.0 for T1 > 2.0 Tc
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Numerical Analysis Assignment NO.2 24-NOV-2010
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-W/g is total mass of building
Distribution of force Fb may based first mode shape and may be by assuminglinear displacement over hight.
Fi =(
,
) * Fb
F4 =(
( )) * Fb = 0.267 * Fb
Q3.3-b. Distribution Seismic force F4 = 1396 KN
-By Upper Limit ( Rigid Diaphram )
Force applied at centre of mass (almost centre of plan) which is
same of centre of area
So, No changes in distribution than in item 2 above
-By Lower Limit ( NO-Diaphram )
Distribution is according to served mass
Each system served mass as same ratio of served wind area.
So, No changes in distribution than in item 2 above