1112012913dynamics response spectrum analysis.pdf

Dynamic Response Spectrum Analysis – Shear Plane Frame of 33

Edited by: Eng.Hussein Rida E-mail: [email protected]

Five-Story Shear Plane Frame Dynamic Response Spectrum Analysis

Comparison between hand calculations based on the theory of structural dynamics and

ETABS analysis procedure results

Problem: Five-story shear plane frame with story-height of 3.0m and single bay of 4.0m

• The mathematical model consists from squares columns ( 26060 cm× ) with infinitely rigid beams ( ∞=beamI ).

• The entire mass of each story is assumed to be lumped at its level with total value of typical story mass ( mkNm /sec.100 2= ).

• The material of columns and beams has modulus of elasticity equal to ( 26 /10.2 mkNE = ).

• Assumed damping ratio ( 05.0=ζ ). • The frame is subjected to dynamic response spectra as defined in UBC-97 with

assumed design parameters : ♦ Seismic zone factor ( 3.0=Z ) ♦ Soil profile type ( BS )

Evaluate 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 to

modal 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 matrix analysis will be done by using MATLAB software (high performance language for technical computing & solve engineering problems).

• Compare hand-calculation results with equivalents obtained from ETABS analysis.



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 the structure 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 of the 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 a structure 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 five degrees 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 structure; the mass matrix is a diagonal matrix (the nonzero elements are only in the main 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

mm

mm

m

M /sec.

10000000100000001000000010000000100

00000000000000000000

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 infinitely rigid comparison to columns; then the story forces can easily be determined by adding the side-sway stiffness of the appropriates stories which equal in this case to the total sum of columns stiffness of that stories. In shear frame as defined previously the stiffness of column with two ends fixed against

rotation is given by: 312

hEIK c

c =

Where ( h ) is the story height, and ( cI ) is the moment of inertia of column's section given by:

433

0108.012

6.06.060.012

mImawhereaaI cc =×

=⇒=×

=

The stiffness of the story is given by:

mkNhEIKKK c

cci /192003

0108.01022424.2 3

6

3 =×××

====∑



The stiffness matrix of the structure is given by:

mkN

KKKKKKK

KKKKKKKK

KK

K /

2100012100

012100012100011

19200

00000

0000000

544

4433

3322

2211

11

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−−

−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+−−+−

−+−−+−

−

=

where iKKandKKKK =54321 ,,, the entire stiffness of the story. • Natural vibration frequencies and corresponding vibration mode shapes:

Based on the dynamics of structures theory, the natural vibration frequencies and corresponding mode shapes can be determined by solve the equation:

0][ 2 =Φ− MK ω This equation is called an eigenvalue problem. The quantities 2ω are the eigenvalues indicating the square of free vibration frequencies, while the corresponding displacement vectors Φ represent the corresponding mode of vibrating system known as the eigenvectors or mode shapes. Hence a nontrivial solution is possible 0≠Φ only when the determinant MK 2ω− equal to zero (due to Cramer's rule). Expanding the determinant will give an algebraic equation of the Nth degree in the frequency parameter 2ω for a system having N degrees of freedom. The N roots of this equation ),...,,,( 22

322

21 Nωωωω 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 mode or the fundamental mode, the next higher frequency is the second mode, etc. It is easily to solve this problem by using MATLAB (Mathematical Programming Language), where mathematically we can write:

))((],[ KMinveig ×=ΩΦ

Where Ω is the vector of square of frequencies.



Type this code in MATLAB editor

MATLAB CODE: >> [ModeShapes,Omega]=eig(inv(M)*K)

The result will be: Square of Frequencies matrix

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=Ω

707.041400000543.5194000003511.329000005335.132000005547.15

To get Frequencies Matrix, type the following code:

MATLAB CODE: >> Freq = zeros(5) >> for i=1:5 Freq(i,i)= omega(i,i)^0.5 end

The result will be:

sec/

26.59020000023.31350000018.14800000011.5123000003.9439

rad

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=ω

Mode shapes matrix:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=Φ

0.03260.05490.05970.0456-0.01700.0549-0.0456-0.01700.0597-0.0326

0.05970.0170-0.0549-0.0326-0.04560.0456-0.05970.0326-0.01700.05490.01700.0326-0.04560.05490.0597

Mode shapes vectors:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=Φ

0.01700.03260.04560.05490.0597

1 ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=Φ

0.0456-0.0597-0.0326-0.01700.0549

2 ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=Φ

0.05970.01700.0549-0.0326-0.0456

3 ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=Φ

0.05490.0456-0.0170-

0.05970.0326-

4 ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=Φ

0.03260.0549-

0.05970.0456-0.0170

5

mode shape-1 mode shape-2 mode shape-3 mode shape-4 mode shape-5

sec/

9439.31

rad=ω

sec/5123.112

rad=ω

sec/1480.183

rad=ω

sec/3135.234

rad=ω

sec/5902.265

rad=ω



The five mode shapes for this frame are sketched below:

sec/9439.31

rad=ω

sec/5123.112

rad=ω

sec/1480.183

rad=ω

sec/3135.234

rad=ω

sec/5902.265

rad=ω

Determination of Period Matrix:

The period (T) of motion is given as a function of frequency as following: (sec)2ωπ

=T

This is mean that each mode shape of vibration has relative period To get the period matrix of the structure; type the following code:

MATLAB CODE: >> Period = zeros(5) >> for i=1:5 Period(i,i) = 2 * pi /freq(i,i) end

The period matrix will be:

sec

0.2363000000.2695000000.3462000000.5458000001.5931

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=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 lesser one is come with second mode shape, etc.



Determination of response spectrum Acceleration Matrix: We can determine the mass acceleration depending on the response spectrum. The response spectrum is a plot of maximum accelerations for all values of periods, in other word for a system has specified period based on its mass and stiffness the response spectrum function gives the maximum acceleration can occur in the entire mass of this system. That is mean; if we know the vibration period of a specific mass we can determine its acceleration depending on response spectrum function. The response spectrum function depending on the site characteristics, therefore the design codes give 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 soil under structure in decrease or increase the amplitude of the motion. So it is very important to know that for a structure has specified period (T) will vibrate in different accelerations due to the site which the structure located. The determination of the design response spectra as per UBC97 requires two design parameters:

⎭⎬⎫

⎩⎨⎧

==

⇒⎭⎬⎫=

3.03.0

)()3(3.0:

V

a

B CC

SprofileSoilZoneforZFactorZoneSeismic

This plot has two characteristics periods

sec08.02.0

sec4.03.05.2

3.05.2

=×=

=×

==

so

a

Vs

TTC

CT



If the period of vibration mode is greater than sT , then the relative acceleration is given by:

2sec/943.281.93.03.0 mTT

gT

gTCS V

a ====

Else, if the period is lesser than sT and greater than oT , then the relative acceleration is given by: 2sec/3575.781.93.05.25.2 mgCS Va =××== To get the acceleration matrix according to the period of vibration mode shapes, type the following code:

MATLAB CODE: >> Sa = zeros(5) >> for i=1:5 if Period(i,i) > 0.4 Sa(i,i) = 0.3 * 9.81 / Period(i,i) else Sa(i,i) = 0.75 * 9.81 end; end

The acceleration matrix will be:

2sec/

7.3575000007.3575000007.3575000005.3923000001.8473

mSa

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

Note: if the structure has a period lesser than or equal to characteristic periods sT , then the entire mass of this structure will vibrate according to the maximum probable acceleration. This will be 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 we have to avoid getting a more flexible structure Determination of maximum modal displacement:

The maximum modal displacement matrix is given by: Ω

Φ= aSmLU *

where: (L) is the matrix of modal excitation factor given by: 1ML TΦ= *m is the generalized modal mass matrix given by: ΦΦ= 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 result will be:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.8853000001.9377000003.4796000006.6022-0000020.9706

L

To get the generalized modal mass matrix, type the following code:

MATLAB CODE: >> ModalMass = ModeShapes' * M * ModeShapes

The result will be:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

1000001000001000001000001

*m

To get the maximum modal displacement, type the following code:

MATLAB CODE: >> U_Modal = ModeShapes * (L/ModalMass) * (Sa/Omega)

The result will be:

mU

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.00030.00140.00460.01220.04230.0005-0.0012-0.00130.01600.08120.00050.0004-0.0043-0.00880.11350.0004-0.00160.0025-0.0046-0.13660.00020.0009-0.00350.0147-0.1487

Where the relative displacement vectors due to each mode shape will be as the following:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.04230.08120.11350.13660.1487

1U ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.01220.01600.00880.0046-0.0147-

2U ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.00460.00130.0043-0.0025-0.0035

3U ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.00140.0012-0.0004-0.00160.0009-

4U ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.00030.0005-0.00050.0004-0.0002

5U

Note that iU is the vector of maximum displacements at story-levels due to relative mode shape iΦ . (we can see clearly that the maximum lateral displacement at 5th story due to 1st, 2nd, 3rd, 4th and 5th mode shapes are given respectively :0.1487, -0.0147, 0.0035, -0.0009, 0.0002 m )



Determination of maximum story-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 individual maxima. 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 a significant 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 and most 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 SRSS approximation of the maximum total displacements is given by:

( ) ( ) ( ) ( ) ( ) ( )252

42

32

22

11

2max UUUUUUU

n

ii ++++== ∑

=

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 together, the SRSS method can 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=1:5 s = s + U_Modal(i,j)^2 end U_Max(i,1) = s^0.5 end

The result will be:

mU

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.04430.08280.11390.13670.1494

max



Determination of maximum modal elastic-forces: The maximum modal elastic forces occur at the story-levels is given by:

as SmLMf *Φ=

To get the matrix of modal elastic forces, type the following code:

MATLAB CODE: >> fs = M * ModeShapes * L /ModalMass * Sa

The result will be:

kNfs

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

21.235978.2015152.8106162.245265.814635.7296-64.9722-43.4944212.4961126.297338.879424.2207-140.4308-116.0654176.548129.6853-85.095583.4652-60.4827-212.4961

11.066246.4792-116.6741195.2809-231.2289

The relative elastic force vectors due to each mode shape will be as the following:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

65.8146126.2973176.5481212.4961231.2289

1sf ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

162.2452212.4961116.0654

60.4827-195.2809-

2sf ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

152.810643.4944140.4308-83.4652-

116.6741

3sf ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

78.201564.9722-24.2207-

85.095546.4792-

4sf ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

21.235935.7296-38.879429.6853-

11.0662

5sf

where sif is the vector of maximum elastic forces at story-levels due to relative mode shape

iΦ . (we can see clearly that the maximum elastic force at 5th story due to 1st, 2nd, 3rd, 4th and 5th mode shapes are given respectively :231.2289, -195.2809, 116.6741, -46.4792, 11.0662 kN) Determinations of maximum modal story shear forces: We can get the shear force acting on a certain story by assembly the elastic-forces acting

above the level of this story. Therefore the story-shear force is given by: ( )∑+=

=n

jiisj fV

1

To assembly the elastic-forces at each story level, type the following code:

MATLAB CODE: >> for i=1:5 for j=1:5 s = 0 for a=1:j s=s+fs(a,i) end V_Modal(j,i)=s end end



The result will be:

kNV

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

5.766727.624989.0831235.0431812.384915.4692-50.5766-63.7274-72.7979746.570320.260414.3956107.2219-139.6982-620.273018.6190-38.616333.2089255.7636-443.724911.066246.4792-116.6741195.2809-231.2289

The relative shear force vectors due to each mode shape will be as the following:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

812.3849746.5703620.2730443.7249231.2289

1V ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

235.043172.7979139.6982-255.7636-195.2809-

2V ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

89.083163.7274-

107.2219-33.2089116.6741

3V ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

27.624950.5766-

14.395638.616346.4792-

4V ,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

5.766715.4692-20.260418.6190-11.0662

5V

where iV is the vector of shear forces at story-levels due to relative mode shape iΦ . (We can see clearly that the shear force at 1st story due to 1st, 2nd, 3rd, 4th and 5th mode shapes are given respectively: 812.3849, 235.0431, 89.0831, 27.6249, 5.7667 kN) Determinations of maximum total story shear forces: Similarly 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 the following:

( ) ( ) ( ) ( ) ( ) ( )252

42

32

22

11

2max VVVVVVV

n

ii ++++== ∑

=

To get the maximum total story shear-forces matrix, type the following code:

MATLAB CODE: >> for i=1:5 s = 0 for j=1:5 s = s+V_Modal(i,j)^2 end V_Max(i,1)=s^0.5 end

The result will be:

kNV

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

850.8506754.6690645.2662515.0219327.8674

max



Determination of modal participation factors: The modal participation factor representing the interaction between the mode shape and the spatial distribution of the external load.

this factor given by: *mLMPF =

To get the matrix of modal participation factor, type the following code:

MATLAB CODE: >> MPF = L / ModalMass

The result will be:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.8853000001.9377000003.4796000006.6022-0000020.9706

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 we need an adequate number of mode shapes to insure that 90% of the mass at least will respond due to earthquake motion. The modal participating mass ratio given by:

( ) %*2

∑=

immLMPMR

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

The result will be:

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

0.1568000000.7509000002.4216000008.71770000087.9530

MPMR



Note that the part of mass which will respond to the motion for 1st, 2nd, 3rd, 4th and 5th are given respectively: 87.95%, 8.72%, 2.42%, 0.75% and 0.16%). Due to MPMR, we see that the first mode shape is most important one, because 87.95% of mass will respond to ground motion, when only 8.72% of mass will respond in the second mode shape, etc. Note that the only first-two mode shapes will be adequate to insure that more than 90% of the mass will vibrate responding to ground motion. References:

Clough, R., and J. Penzien. 1993. Dynamics of Structures, Second Edition. McGraw- Hill.

Paz, M. 1985. Structural Dynamics, theory and computation. Van Nostrand Reinhold.

International Conference of Building Official 1997, Uniform Building Code. Whittier, California.

CSI Analysis Reference Manual. CSI Computers & Structures, Berkeley, California.

ETABS Software Verification Examples, CSI Computers & Structures, Berkeley, California.



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 0 0 100 0 0 0 0 0 100 0 0 0 0 0 100 0 0 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 0 0 0 -19200 38400 -19200 0 0 0 -19200 38400 >> % ModeShapes & Squared Frequencies >> [ModeShapes,Omega]=eig(inv(M)*K) Omega = 15.5547 0 0 0 0 0 132.5335 0 0 0 0 0 329.3511 0 0 0 0 0 543.5194 0 0 0 0 0 707.0414 ModeShapes = 0.0597 0.0549 0.0456 -0.0326 0.0170 0.0549 0.0170 -0.0326 0.0597 -0.0456 0.0456 -0.0326 -0.0549 -0.0170 0.0597 0.0326 -0.0597 0.0170 -0.0456 -0.0549 0.0170 -0.0456 0.0597 0.0549 0.0326



MATLAB CODE >> Freq = zeros(5) >> for i=1:5 Freq(i,i)= omega(i,i)^0.5 end Freq = 3.9439 0 0 0 0 0 11.5123 0 0 0 0 0 18.1480 0 0 0 0 0 23.3135 0 0 0 0 0 26.5902 >> %Period Matrix >> Period = zeros(5) >> for i=1:5 Period(i,i) = 2 * pi /freq(i,i) end Period = 1.5931 0 0 0 0 0 0.5458 0 0 0 0 0 0.3462 0 0 0 0 0 0.2695 0 0 0 0 0 0.2363 >> %Acceleration Matrix >> Sa = zeros(5) >> for i=1:5 if Period(i,i) > 0.4 Sa(i,i) = 0.3 * 9.81 / Period(i,i) else Sa(i,i) = 0.75 * 9.81 end; end Sa = 1.8473 0 0 0 0 0 5.3923 0 0 0 0 0 7.3575 0 0 0 0 0 7.3575 0 0 0 0 0 7.3575



MATLAB CODE >> %Modal Excitation Matrix >> LL = ModeShapes' * M * [1;1;1;1;1] >> for i=1:5 L(i,i) = LL(i,1) end L = 20.9706 0 0 0 0 0 -6.6022 0 0 0 0 0 3.4796 0 0 0 0 0 1.9377 0 0 0 0 0 0.8853 >> % Modal Mass Matrix >> ModalMass = ModeShapes' * M * ModeShapes ModalMass = 1.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 0.0000 0.0000 0.0000 1.0000 0.0000 0.0000 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 0.0002 0.1366 -0.0046 -0.0025 0.0016 -0.0004 0.1135 0.0088 -0.0043 -0.0004 0.0005 0.0812 0.0160 0.0013 -0.0012 -0.0005 0.0423 0.0122 0.0046 0.0014 0.0003



MATLAB CODE >> %Maximum Story Displacement due to SRSS Combination >> for i=1:5 s = 0 for j=1:5 s = s + U_Modal(i,j)^2 end U_Max(i,1) = s^0.5 end U_Max = 0.1494 0.1367 0.1139 0.0828 0.0443 >> %Maximum Modal Elastic Forces Matrix >> fs = M * ModeShapes * L /ModalMass * Sa fs = 231.2289 -195.2809 116.6741 -46.4792 11.0662 212.4961 -60.4827 -83.4652 85.0955 -29.6853 176.5481 116.0654 -140.4308 -24.2207 38.8794 126.2973 212.4961 43.4944 -64.9722 -35.7296 65.8146 162.2452 152.8106 78.2015 21.2359 >> %Maximum Modal Story-Shear Matrix >> for i=1:5 for j=1:5 s = 0 for a=1:j s=s+fs(a,i) end V_Modal(j,i)=s end end V_Modal = 231.2289 -195.2809 116.6741 -46.4792 11.0662 443.7249 -255.7636 33.2089 38.6163 -18.6190 620.2730 -139.6982 -107.2219 14.3956 20.2604 746.5703 72.7979 -63.7274 -50.5766 -15.4692 812.3849 235.0431 89.0831 27.6249 5.7667



MATLAB CODE >> %Maximum Total Story Shear Forces due to SRSS Combination >> for i=1:5 s = 0 for j=1:5 s = s+V_Modal(i,j)^2 end V_Max(i,1)=s^0.5 end V_Max = 327.8674 515.0219 645.2662 754.6690 850.8506 >> %Modal Participation Factor >> MPF = L / ModalMass MPF = 20.9706 0.0000 0.0000 0.0000 0.0000 0.0000 -6.6022 0.0000 0.0000 0.0000 0.0000 0.0000 3.4796 0.0000 0.0000 0.0000 0.0000 0.0000 1.9377 0.0000 0.0000 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 0.0000 0.0000 8.7177 0.0000 0.0000 0.0000 0.0000 0.0000 2.4216 0.0000 0.0000 0.0000 0.0000 0.0000 0.7509 0.0000 0.0000 0.0000 0.0000 0.0000 0.1568



ETABS MODEL

Build of Mathematical Computer Model: • The frame is modeled as five-story consist from two-column line, singly bay system with

story-height 3m & length of bay 4m. kN-m-second units are used.



ETABS MODEL • Define material properties: (Modulus of elasticity, Self-Mass of Material)

Assume that the self-weight of the frame elements is neglected



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.



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.



ETABS MODEL

• Build the model to be as following:



ETABS MODEL • Draw Point Object at the mid-span of beams in order to assign lumped mass at the story-

level.



ETABS MODEL

• Assign Lumped Mass at story-level.



ETABS MODEL

• Define Mass Source:

• Assign Diaphragm at Story-Level:



ETABS MODEL

• Define Response Spectrum Function (UBC97 Design Spectrum):



ETABS MODEL

• Define & Assign Response Spectrum Case Data:



ETABS MODEL

• Define Analysis options:

• Perform analysis.



ETABS MODEL • Building Mode-Shapes:



ETABS MODEL

• Mode-Shapes:



ETABS MODEL

• Periods and Accelerations:

• Maximum Story-Displacement according to SRSS combination:

• Maximum Story-Shear Force according to SRSS combination:



ETABS MODEL

• Modal Participation Factor:

• Modal Participating Mass Ratio:

1112012913dynamics response spectrum analysis.pdf

Documents