dynamics response spectrum analysis - shear plane frame

8/13/2019 Dynamics Response Spectrum Analysis - Shear Plane Frame

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Dynamic Response Spectrum Modal Analysis Shear Plane Frame Page 1 of 35

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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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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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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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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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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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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ETABS MODEL

Assign Lumped Mass at story-level.


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ETABS MODEL

Define Mass Source:

Assign Diaphragm at Story-Level:


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ETABS MODEL

Define Response Spectrum Function (UBC97 Design Spectrum):


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ETABS MODEL

Define & Assign Response Spectrum Case Data:


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ETABS MODEL

Define Analysis options:

Perform analysis.


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ETABS MODEL

Building Mode-Shapes:


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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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dynamics response spectrum analysis - shear plane frame

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