Dynamics Response Spectrum Analysis - Shear Plane Frame
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Transcript of Dynamics Response Spectrum Analysis - Shear Plane Frame
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Edited by: Eng. Hussein RidaE-mail: [email protected]
Five-Story Shear Plane Frame
Dynamic Response Spectrum Modal Analysis
Problem:
Shear plane frame consists of five typical stories of 3m height and 4m bay as shown on belowfigure with following characteristics:
1. The mathematical model consists of squares columns ( 26060 cm ) with infinitely rigidbeams ( beamI ).
2. The entire mass of each story is assumed to be lumped at its level with total value of mass( mkNm /sec.100 2 ).
3. The material of columns and beams has modulus of elasticity equal to( 26 /10.2 mkNE ).
4. Assumed damping ratio ( 05.0 ).5. The frame is subjected to ground spectral response acceleration as defined in UBC-97,
figure16-3 with following parameters:
Seismic zone factor ( 3.0Z ) Soil profile type (
BS )
For the given frame determine the following:(a).Natural vibration frequencies and corresponding vibration
mode shapes.
(b).Periods corresponding to vibration mode shapes.(c).Response spectrum accelerations corresponding to periods.(d).Maximum modal displacement corresponding to
vibration mode shapes.
(e).Maximum story-displacement according tomodal combination (SRSS).(f).Maximum modal elastic forces (inertia-forces) at story-levels.
(g).Maximum modal story-shear forces.(h).Maximum total story-shear forces according to
modal combination (SRSS).
(i).Modal participation factors.(j).Modal participating mass ratios.
Notes:
The dynamic modal analysis will be carried out in two different ways:1. Utilizing MATLAB to perform mathematical matrix analysis based on theory of structural
dynamics.2. Utilizing ETABS to perform finite element analysis along with modal analysis.
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Introduction:A shear frame may be defined as a structure in which there is no rotation of a horizontal
section at the level of the floor. In this respect the deflected frame will have many of the
features of a cantilever beam that is deflected by shear forces, Hence the name Shear Frame.
To accomplish such deflection in frame, we must assume that: (1) the total mass of thestructure is concentrated at the levels of the floors; (2) the beams on the floor are infinitely
rigid as compared to the columns; and (3) the deformation of the structure is independent ofthe axial forces present in the columns. These assumptions transform the problem from a
structure with an infinite number of degree of freedom (due to the distributed mass) to astructure which has only as many degrees as it has lumped masses at the floor levels.
According to previous discussion a five stories frame modeled as a shear frame will have fivedegrees of freedom, that is, the five horizontal displacements at the floor levels. The second
assumption introduces the requirement that the joints between beams and columns are fixed
against rotation. The third assumption leads to the condition that the rigid beams will remain
horizontal during motion.
Determination of Lumped mass matrix:
For shear frame structure; the mass matrix is a diagonal one with nonzero elements located at
its diagonal whereas each one of these elements represents the total equivalent entire mass of
the story as a concentrated lumped mass at the level of this story with understanding that only
horizontal displacement of this mass is possible.
Therefore the lumped mass matrix is given by:
mkN
m
m
m
m
m
M /sec.
1000000
0100000
0010000
0001000
0000100
0000
0000
0000
0000
0000
2
5
4
3
2
1
Determination of stiffness matrix:The stiffness matrix of shear frame can be determined by applying a unit displacement to each
story alternately and evaluation the resulting story forces. Because the beams are infinitelyrigid comparison to columns; then the story forces can easily be determined by adding the
side-sway stiffness of the appropriate stories which equal in this case to the total sum ofcolumns stiffness of those stories.
For shear frame as defined previously the stiffness of column with two ends fixed against
rotation is given by:3
12
h
EIK c
c
Where ( h ) is the story height, and ( cI ) is the moment of inertia of column's section given by:
433
0108.012
6.06.060.0
12mImawhere
aaI cc
The stiffness of the story is given by:
mkNh
EIKKK ccci /19200
3
0108.01022424.2
3
6
3
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The stiffness matrix of the structure is given by:
mkN
KKK
KKKK
KKKK
KKKK
KK
K /
21000
12100
01210
00121
00011
19200
000
00
00
00
000
544
4433
3322
2211
11
Where iKKandKKKK 54321 ,,, are the entire stiffness of each story respectively.
Natural vibration frequencies and corresponding vibration mode shapes:Based on the dynamics of structures theory, the natural vibration frequencies andcorresponding mode shapes can be determined by solving the equation:
0][ 2 MK
This equation is called an eigenvalue problem. The quantities 2 are the eigenvalues
indicating the square of free vibration frequencies, while the corresponding displacement
vectors represent the corresponding mode of vibrating system known as the eigenvectorsor mode shapes.
Hence a nontrivial solution is possible 0 only when the determinant MK2 equal
to zero per Cramer's rule. Expanding the determinant will give an algebraic equation of the
Nth
degree in the frequency parameter 2 for a system having N degrees of freedom. The
N roots of this equation ),...,,,(22
3
2
2
2
1 N represent the frequencies of the N modes of
vibration which are possible in the system. The mode having the lowest frequency is called
the first or fundamental mode shape, the next bigger frequency is called the second modeshape, and so on.
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Above Eigen value problem can be expressed as follow:
))((],[ KMinveig
Where is the vector of square of frequencies, and is the matrix of mode shapes.
To solve this problem, type the following code in MATLAB text editor.
MATLAB CODE:
>> [ModeShapes,Omega]=eig(inv(M)*K)
The output results:
Square of Frequencies matrix
707.04140000
0543.5194000
003511.32900
0005335.1320
00005547.15
To get Frequencies Matrix, type the following code:
MATLAB CODE:
>> Freq = zeros(5)>> for i=1:5
Freq(i,i)= omega(i,i)^0.5end
The output result:
sec/
26.59020000
023.3135000
0018.148000
00011.51230
00003.9439
rad
Mode shapes matrix:
0.03260.05490.05970.0456-0.0170
0.0549-0.0456-0.01700.0597-0.03260.05970.0170-0.0549-0.0326-0.0456
0.0456-0.05970.0326-0.01700.0549
0.01700.0326-0.04560.05490.0597
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Mode shapes vectors:
0.0170
0.03260.0456
0.0549
0.0597
1 ,
0.0456-
0.0597-0.0326-
0.0170
0.0549
2 ,
0.0597
0.01700.0549-
0.0326-
0.0456
3 ,
0.0549
0.0456-0.0170-
0.0597
0.0326-
4 ,
0.0326
0.0549-0.0597
0.0456-
0.0170
5
mode shape-1 mode shape-2 mode shape-3 mode shape-4 mode shape-5
sec/
9439.31
rad
sec/
5123.112
rad
sec/
1480.183
rad
sec/
3135.234
rad
sec/
5902.265
rad
The five mode shapes for this frame are sketched below:
sec/
9439.31
rad
sec/
5123.112
rad
sec/
1480.183
rad
sec/
3135.234
rad
sec/
5902.265
rad
Determination of Period Matrix:
The period (T) of motion is given as a function of frequency as follow: (sec)2
T
This means that each mode shape of vibration has its relative period
To get theperiod matrixof the structure; type the following code:
MATLAB CODE:>> Period = zeros(5)>> for i=1:5
Period(i,i) = 2 * pi /freq(i,i)end
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The period matrix:
sec
0.23630000
00.2695000
000.346200
0000.54580
00001.5931
T
Where the period of 1st
, 2nd
, 3rd
, 4th
and 5th
mode shapes are given respectively: (1.5931,
0.5458, 0.3462, 0.2695, 0.2363 sec).
Note that the period of the first mode shape is the biggest one (T=1.5931 sec) which is called
the fundamental period. The next bigger one is come with second mode shape, and so on.
Determination of response spectrum Acceleration Matrix:
Mass acceleration is determined based on response spectrum function.
The response spectrum is a plot of maximum accelerations for different period values, in
other word; for a system has specific period based on its mass and stiffness, the responsespectrum function gives the maximum acceleration can occur in this mass. This means; if the
period of a specific mass is known, so the mass-acceleration can be determined based on
response spectrum function.
The response spectrum function depends on site characteristics, therefore the design codesgive the response spectrum as a function of zone and soil profile, where the zone reflects the
acceleration occur in the mother bed rock, and the soil profile reflect the effect of the soilunder structure in decreasing or increasing the amplitude of the motion. So it is very
important to know that for a structure has specified period (T), so it will vibrate in differentaccelerations due to the site where the structure founded on.
The determination of the design response spectra as per UBC97 requires two designparameters:
3.0
3.0
)(
)3(3.0:
V
a
B C
C
SprofileSoil
ZoneforZFactorZoneSeismic
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This plot has two characteristics periods
sec08.02.0
sec4.03.05.2
3.0
5.2
so
a
Vs
TT
C
CT
If the period of vibration mode is greater than sT , then the relative acceleration is given by:
2sec/943.2
81.93.03.0
mTT
gT
gT
CS Va
Else, if the period is lesser than sT and greater than oT , then the relative acceleration is given
by: 2sec/3575.781.93.05.2.5.2 mgCS aa
Following MATALB code gives the acceleration matrix according to the period of vibration
mode shapes:
MATLAB CODE:>> Sa = zeros(5)>> for i=1:5
if Period(i,i) > 0.4Sa(i,i) = 0.3 * 9.81 / Period(i,i)
elseSa(i,i) = 0.75 * 9.81
end;end
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The acceleration matrix:
2sec/
7.35750000
07.3575000
007.357500
0005.39230
00001.8473
mSa
Note: if the structure has a period lesser than or equal to characteristic periodssT, then the
entire mass of this structure will be excited according to the maximum probable acceleration.
This will lead to create a maximum inertia force in mass. Therefore it is very important to
scale the ratio of the structure's stiffness to its mass to get a value of period more than sT as
much as possible, but at the same time special attention shall be paid to avoid getting a
flexible structure
Determination of maximum modal displacement:The maximum modal displacement matrix is given by:
a
S
m
LU
*
where: (L) is the matrix of modal excitation factor given by: 1ML T ( *m ) is the generalized modal mass matrix given by: Mm T*
To get the matrix of modal excitation factor; type the following code:
MATLAB CODE:>> LL = ModeShapes' * M * [1;1;1;1;1]>> for i=1:5
L(i,i) = LL(i,1)end
The output result:
0.88530000
01.9377000
003.479600
0006.6022-0
000020.9706
L
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Determination of maximumstory-displacement:
Maximum total response cannot be obtained, in general, by merely adding the modal maxima
because these maxima usually do not occur at the same time. In most cases, when one mode
achieves its maximum response, the other modal responses are less than their individualmaxima. Therefore, although the superposition of the modal spectral values obviously
provides an upper limit to the total response, it generally over estimates this maximum by asignificant amount. A number of different formulas have been proposed to obtain a more
reasonable estimate of the maximum response from the spectral values. The simplest andmost popular of these is the square root of the sum of the squares (SRSS) of the maximum
modal responses. Thus if the maximum modal displacements are given as previous, the SRSSapproximation of the maximum total displacements is given by:
25
2
4
2
3
2
2
2
1
1
2
max UUUUUUUn
i
i
Where the terms under the radical sign represent the vectors of the maximum modal
displacements squared.
It is very important to know that the SRSS method is fundamentally sound when the modal
frequencies are well separated. However, when the frequencies of major contributing modes
are very close to each other, then SRSS method may give poor results, in which case the more
general complete quadratic combination (CQC) method should be used.
To get the maximum total displacement matrix, type the following code:
MATLAB CODE:>> for i=1:5
s = 0
for j=1:5s = s + U_Modal(i,j)^2
endU_Max(i,1) = s^0.5
end
The output result:
mU
0.0443
0.0828
0.1139
0.1367
0.1494
max
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Determination of maximum modal elastic-forces:
The maximum modal elastic forces occur at the story-levels is given by:
asS
m
LMf
*
To get the matrix of modal elastic forces, type the following code:
MATLAB CODE:>> fs = M * ModeShapes * L /ModalMass * Sa
The output result:
kNfs
21.235978.2015152.8106162.245265.8146
35.7296-64.9722-43.4944212.4961126.2973
38.879424.2207-140.4308-116.0654176.5481
29.6853-85.095583.4652-60.4827-212.4961
11.066246.4792-116.6741195.2809-231.2289
The relative elastic force vectors due to each mode shape are:
65.8146
126.2973
176.5481
212.4961
231.2289
1sf ,
162.2452
212.4961
116.0654
60.4827-
195.2809-
2sf ,
152.8106
43.4944
140.4308-
83.4652-
116.6741
3sf ,
78.2015
64.9722-
24.2207-
85.0955
46.4792-
4sf ,
21.2359
35.7296-
38.8794
29.6853-
11.0662
5sf
Wheresif is the vector of maximum elastic forces at story-levels due to relative mode
shape i .
(It is obviously shown that the maximum elastic force at top 5 thstory due to 1st, 2 nd, 3 rd, 4th
and 5th
mode shapes are given respectively :231.2289, -195.2809, 116.6741, -46.4792,
11.0662 kN)
Determination of maximum modal story shear force:
Shear force acting on certain story level is determined by assemble the elastic-forces acting
above the level of this story. Therefore the story-shear force is given by:
n
ji
isj fV1
To assemble the elastic-forces at each story level, type the following code:
MATLAB CODE:
>> for i=1:5for j=1:5s = 0for a=1:j
s=s+fs(a,i)end
V_Modal(j,i)=send
end
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The output result:
kNV
5.766727.624989.0831235.0431812.3849
15.4692-50.5766-63.7274-72.7979746.5703
20.260414.3956107.2219-139.6982-620.2730
18.6190-38.616333.2089255.7636-443.7249
11.066246.4792-116.6741195.2809-231.2289
The relative shear force vectors due to each mode shape are:
812.3849
746.5703
620.2730
443.7249
231.2289
1V ,
235.0431
72.7979
139.6982-
255.7636-
195.2809-
2V ,
89.0831
63.7274-
107.2219-
33.2089
116.6741
3V ,
27.6249
50.5766-
14.3956
38.6163
46.4792-
4V ,
5.7667
15.4692-
20.2604
18.6190-
11.0662
5V
Where iVis the vector of shear forces at story-levels due to relative mode shape i .
(it is obviously shown that the shear force at 1st story due to 1
st, 2
nd, 3
rd, 4
th and 5
thmode
shapes are given respectively: 812.3849, 235.0431, 89.0831, 27.6249, 5.7667 kN)
Determination of maximum total story shear force:
Similar to previous (Determination of maximum story-level displacement), the maximum
total story shear forces could be approximated from the modal maxima by using SRSS
combination method as following:
2
5
2
4
2
3
2
2
2
1
1
2
max VVVVVVV
n
i
i To get the maximum total story shear-forces matrix, type the following code:
MATLAB CODE:>> for i=1:5
s = 0for j=1:5s = s+V_Modal(i,j)^2
endV_Max(i,1)=s^0.5
end
The output result:
kNV
850.8506
754.6690
645.2662
515.0219
327.8674
max
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Determination of modal participation factors:
The modal participation factor represents the interaction between the mode shape and the
spatial distribution of the external load.
This factor is given by:*m
LMPF
To get the matrix of modal participation factor, type the following code:
MATLAB CODE:>> MPF = L / ModalMass
The output result:
0.88530000
01.9377000
003.479600
0006.6022-0
000020.9706
MPF
Determination of modal participating mass ratio:
The modal participating mass ratio represent the part of the total mass which responding to
earthquake motion in each mode, therefore this ratio is very important to determine the
adequate number of mode shapes which give a reasonable part of vibration mass which will
respond to the motion.
The UBC-97 Code declares that adequate numbers of mode shapes are needed to insure that
90% of mass at least will respond due to earthquake motion.
The modal participating mass ratio given by:
%
*2
im
mLMPMR
To get the modal participating mass ratio matrix, type the following code:
MATLAB CODE:>> Segma_M = 0>> for i=1:5
Segma_M = Segma_M + M(i,i)
end>> MPMR = ((L*L/ModalMass)/Segma_M)*100
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The output result:
0.15680000
00.7509000
002.421600
0008.71770
000087.9530
MPMR
Note that the portion of entire structure mass which will respond to the motion for 1st, 2
nd, 3
rd,
4th
and 5th
are given respectively: 87.95%, 8.72%, 2.42%, 0.75% and 0.16%).
According to MPMR, it is obviously shown that the first mode shape is the most important
one, since 87.95% of entire mass will respond to ground motion, when only 8.72% of mass
will respond in the second mode shape, and so on.
Note that only the first-two mode shapes are adequate to insure that more than 90% of the
entire mass will vibrate responding to ground motion.
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MATLAB CODE
>> % Define Lumped Mass Matrix>> M = [100 0 0 0 0;
0 100 0 0 0;
0 0 100 0 0;0 0 0 100 0;0 0 0 0 100]
M =
100 0 0 0 00 100 0 0 00 0 100 0 00 0 0 100 00 0 0 0 100
>> % Define Stiffness Matrix
>> K = 19200.*[1 -1 0 0 0;-1 2 -1 0 0;0 -1 2 -1 0;0 0 -1 2 -1;0 0 0 -1 2]
K =
19200 -19200 0 0 0-19200 38400 -19200 0 0
0 -19200 38400 -19200 00 0 -19200 38400 -19200
0 0 0 -19200 38400
>> % ModeShapes & Squared Frequencies>> [ModeShapes,Omega]=eig(inv(M)*K)
Omega =
15.5547 0 0 0 00 132.5335 0 0 00 0 329.3511 0 00 0 0 543.5194 00 0 0 0 707.0414
ModeShapes =
0.0597 0.0549 0.0456 -0.0326 0.01700.0549 0.0170 -0.0326 0.0597 -0.04560.0456 -0.0326 -0.0549 -0.0170 0.05970.0326 -0.0597 0.0170 -0.0456 -0.05490.0170 -0.0456 0.0597 0.0549 0.0326
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MATLAB CODE
>> Freq = zeros(5)>> for i=1:5
Freq(i,i)= omega(i,i)^0.5
end
Freq =
3.9439 0 0 0 00 11.5123 0 0 00 0 18.1480 0 00 0 0 23.3135 00 0 0 0 26.5902
>> %Period Matrix>> Period = zeros(5)
>> for i=1:5Period(i,i) = 2 * pi /freq(i,i)end
Period =
1.5931 0 0 0 00 0.5458 0 0 00 0 0.3462 0 00 0 0 0.2695 00 0 0 0 0.2363
>> %Acceleration Matrix>> Sa = zeros(5)>> for i=1:5
if Period(i,i) > 0.4Sa(i,i) = 0.3 * 9.81 / Period(i,i)
elseSa(i,i) = 0.75 * 9.81
end;end
Sa =
1.8473 0 0 0 00 5.3923 0 0 00 0 7.3575 0 00 0 0 7.3575 00 0 0 0 7.3575
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MATLAB CODE
>> %Modal Excitation Matrix>> LL = ModeShapes' * M * [1;1;1;1;1]
>> for i=1:5L(i,i) = LL(i,1)end
L =
20.9706 0 0 0 00 -6.6022 0 0 00 0 3.4796 0 00 0 0 1.9377 00 0 0 0 0.8853
>> % Modal Mass Matrix>> ModalMass = ModeShapes' * M * ModeShapes
ModalMass =
1.0000 0.0000 0.0000 0.0000 0.00000.0000 1.0000 0.0000 0.0000 0.00000.0000 0.0000 1.0000 0.0000 0.00000.0000 0.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 0.0000 1.0000
>> %Modal Displacement>> U_Modal = ModeShapes * (L/ModalMass) * (Sa/Omega)
U_Modal =
0.1487 -0.0147 0.0035 -0.0009 0.00020.1366 -0.0046 -0.0025 0.0016 -0.00040.1135 0.0088 -0.0043 -0.0004 0.00050.0812 0.0160 0.0013 -0.0012 -0.0005
0.0423 0.0122 0.0046 0.0014 0.0003
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MATLAB CODE
>> %Maximum Story Displacement due to SRSS Combination>> for i=1:5
s = 0
for j=1:5s = s + U_Modal(i,j)^2end
U_Max(i,1) = s^0.5end
U_Max =
0.14940.13670.11390.0828
0.0443
>> %Maximum Modal Elastic Forces Matrix>> fs = M * ModeShapes * L /ModalMass * Sa
fs =
231.2289 -195.2809 116.6741 -46.4792 11.0662212.4961 -60.4827 -83.4652 85.0955 -29.6853176.5481 116.0654 -140.4308 -24.2207 38.8794126.2973 212.4961 43.4944 -64.9722 -35.729665.8146 162.2452 152.8106 78.2015 21.2359
>> %Maximum Modal Story-Shear Matrix>> for i=1:5
for j=1:5s = 0for a=1:js=s+fs(a,i)
endV_Modal(j,i)=s
endend
V_Modal =
231.2289 -195.2809 116.6741 -46.4792 11.0662443.7249 -255.7636 33.2089 38.6163 -18.6190620.2730 -139.6982 -107.2219 14.3956 20.2604746.5703 72.7979 -63.7274 -50.5766 -15.4692812.3849 235.0431 89.0831 27.6249 5.7667
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MATLAB CODE
>> %Maximum Total Story Shear Forces due to SRSS Combination>> for i=1:5
s = 0
for j=1:5s = s+V_Modal(i,j)^2end
V_Max(i,1)=s^0.5end
V_Max =
327.8674515.0219645.2662754.6690
850.8506
>> %Modal Participation Factor>> MPF = L / ModalMass
MPF =
20.9706 0.0000 0.0000 0.0000 0.00000.0000 -6.6022 0.0000 0.0000 0.00000.0000 0.0000 3.4796 0.0000 0.00000.0000 0.0000 0.0000 1.9377 0.00000.0000 0.0000 0.0000 0.0000 0.8853
>> %Modal Participating Mass Ratio>> Segma_M = 0>> for i=1:5
Segma_M = Segma_M + M(i,i)end
>> MPMR = ((L*L/ModalMass)/Segma_M)*100
MPMR =
87.9530 0.0000 0.0000 0.0000 0.00000.0000 8.7177 0.0000 0.0000 0.0000
0.0000 0.0000 2.4216 0.0000 0.00000.0000 0.0000 0.0000 0.7509 0.00000.0000 0.0000 0.0000 0.0000 0.1568
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Build of Mathematical Computer Model:
The frame is modeled as five-story consist from two-column line, singly bay system withstory-height 3m & length of bay 4m. kN-m-second units are used.
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Define material properties: (Modulus of elasticity, Self-Mass of Material)Assume that the self-weight of the frame elements is neglected
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Define Column's Properties:The column is modeled to have infinite axial area, so that axial deformation is neglected.Also, Zero column shear area is input to trigger the ETABS option of neglecting shear
deformation. These deformations are neglected to be consistent with the hand-calculated
model with which the result are compared.
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Define Beam's Properties:The beam is modeled as a rigid beam to have infinite moment of inertia compared to column,so that axial deformation is neglected. Also, neglecting both shear deformations and axial
deformations.
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Draw Point Object at the mid-span of beams in order to assign lumped mass at the story-level.
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Assign Lumped Mass at story-level.
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Define Mass Source:
Assign Diaphragm at Story-Level:
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Define Response Spectrum Function (UBC97 Design Spectrum):
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Define & Assign Response Spectrum Case Data:
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Define Analysis options:
Perform analysis.
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Building Mode-Shapes:
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Edited by: Eng. Hussein RidaE-mail: [email protected]
ETABS MODEL
Mode-Shapes:
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ETABS MODEL
Modal Participation Factor:
Modal Participating Mass Ratio:
Conclusion:It is obviously shown that the output results of ETABS are almost identical to those obtained by
MATLAB developed algorithm based on theory of structural dynamics.
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