Linear Modal Time History Analysis V1.1

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Linear Time History Analysis of MDOF Structure by Mode Superposition Method using Newmark’s β Method Carleton University © Mostafa Tazarv Graduate Student Version 1.1 (April 2011) Cite as: Tazarv, M., “Linear Time History Analysis of MDOF Structures by Mode Superposition Method”, available at: http://alum.sharif.edu/~tazarv/

Transcript of Linear Modal Time History Analysis V1.1

Page 1: Linear Modal Time History Analysis V1.1

Linear Time History Analysis

of MDOF Structure by Mode

Superposition Method using Newmark’s β Method

Carleton University

© Mostafa Tazarv

Graduate Student

Version 1.1 (April 2011)

Cite as: Tazarv, M., “Linear Time History Analysis of MDOF Structures by

Mode Superposition Method”, available at: http://alum.sharif.edu/~tazarv/

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Introduction

Structural dynamic is a mandatory graduate level course for structural/earthquake engineering

student all around the world. One of the most important topics of this course is to solve modal

equations of motion of a Multi Degree of Freedom (MDOF) structure by a numerical method

such as Nemark’s β method and Wilson-Ѳ method. Modal response should be assembled to

obtain each DOF response. Applied load can be a base excitation earthquake or time-dependent

loads on stories.

Here, I will introduce a MATLAB function which can do a time history analysis of an n-DOF

structure with a certain Number of Modes (nom). Then, I will show how to use this function with

two examples, one excited by half-cycle harmonic on two stories and another excited by Elcentro

earthquake. This function will also be verified by an example detailed in “Dynamic of Structure”

by Chopra.

Time History Response by Newmark Method: NM

I tried to write a self-explanatory m-file. Therefore, I’ll explain a bit more the important part of

the function called “NM” in this section. A new feature is to give you the option to specify the

number of modes you want to consider in mode superposition analysis. For example, there is a

100-DOF structure (the size of mass and stiffness matrixes are 100�100). However, you want to

do the modal analysis only for first 10 modes not all the modes which is very common in real

situation.

In this function to solve equation of motions for different modes, Newmark Linear Method has

been used. We can decompose time-dependent applied force, ���� � �. ���, into two

components where F is spatial distribution of load on DOFs and f(t) is time-variant component of

load. By modifying inputs F (a vector) and f(t), you can analyze the structure for either

earthquake (base-seismic-excitation) or time-dependent load applied to different stories. I will

show it in examples later. That's your responsibility to organize eigenvectors (Φ) and

eigenvaluses (�� in which frequencies are sorted from smallest to greatest ( f1< f2< f3<…). If

you want to use MATLAB built-in function to find Φ (phi) and �(omega2), you should be

careful about abovementioned arrangement.

Preferred unit is SI (N, kg, s). Specifically for this function, stiffness (K) is in N/m, mass (M) is

in kg, force (F) is in N, Gravitational Constant (g) is 9.810 m/s2 and Ground Acceleration (Ag)

should be normalized to g. However, you are free to choose your unit.

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function [disp,EF]=NM(n,nom,dt,F,ft,M,K,zet,omega2,phi)

% INPUTS:% INPUTS:% INPUTS:% INPUTS:

% nnnn: Number of Stories or generally, Number fo DOFs

% nomnomnomnom: Number of Modes that you want to consider in analysis; (nom <= n)

% dtdtdtdt: Time Interval (Sampling Time) of force or earthquake

% FFFF: Spatial Load Distribution Vector; in the case of base seismic

% exitation F=-g*M*[1 1 1 ... 1]'; (A vector)

% ftftftft: Time-Variant Component of Load (A Vector); in the

% case of base seismic exitation ft=Ag (Ground Acceleration in g)

% MMMM: Mass Matrix of Stories (n x n)

% KKKK: Stiffness Matrix of Stories (n x n)

% zetzetzetzet: Damping Ratio in percent (%); e.g. 5

% omega2omega2omega2omega2: Eigen-Values Matrix (nom x nom); e.g. [phi,omega2]=eig(K,M);

% phiphiphiphi: Eigen-Vectors (nom x nom); e.g. [phi,omega2]=eig(K,M);

% OUTPUTS (Story Response):% OUTPUTS (Story Response):% OUTPUTS (Story Response):% OUTPUTS (Story Response):

% dispdispdispdisp: Displacement Time History of Stories (or DOFs)

% EFEFEFEF: Elastic Force Time History of Stories (or DOFs)

As long as the size of stiffness matrix and mass matrix are equal, there will be no restriction on this

function.

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Example 1: Time-Dependent Load on a Shear Building with 8-DOF

It is desired to analyze a shear building detailed in Figure 1 with Mode Superposition Method

only with first 2 modes in an undamped condition (zet=0). Load considered for this example is

time-dependent half-sine impulse force on two stories (Figure 3). For a shear frame, it is easy to

derive stiffness and mass matrixes which are shown as follows:

Figure 1- Shear Building

−+−

−+

=

nk

k

kkkk

kkk

K

...00

......0

0

00

3

3322

221

,

=

nm

m

m

M

...00

......00

000

000

2

1

)/(

0.5 0.5- 0 0 0 0 0 0

0.5- 1.0 0.5- 0 0 0 0 0

0 0.5- 1.0 0.5- 0 0 0 0

0 0 0.5- 1.0 0.5- 0 0 0

0 0 0 0.5- 1.0 0.5- 0 0

0 0 0 0 0.5- 1.0 0.50- 0

0 0 0 0 0 0.5- 1.5 1.0-

0 0 0 0 0 0 1.0- 2.0

109 mNK

×=

)/.(

4 0 0 0 0 0 0 0

0 4 0 0 0 0 0 0

0 0 4 0 0 0 0 0

0 0 0 8 0 0 0 0

0 0 0 0 8 0 0 0

0 0 0 0 0 8 0 0

0 0 0 0 0 0 8 0

0 0 0 0 0 0 0 8

10 26 msNkgM =

×=

Derived mass and stiffness matrixes can be used as inputs of eigen-problem and modal analysis.

For this section, only 2 modes are desired. Then, eigenvalues and eigenvectors have been

calculated. Figure 2 illustrates modal shape of first two modes normalized to mass.

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Figure 2- Mode Shapes of First Two Modes of Shear Building (Normalized to Mass)

×=Φ −

0.2429 0.2132-

0.1849 0.2062-

0.0828 0.1923-

0.0391- 0.1720-

0.1423- 0.1403-

0.1776- 0.0994-

0.1280- 0.0518-

0.0727- 0.0264-

103

� � �4.1408 00 29.8429�

NOTE: � should be a square matrix in the size of nom � nom

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As mentioned before, to solve equation of motion of each mode,

n

T

nnnnnnnMtfFYYY /)(..2 2 ϕωωξ =++ &&& , load should be in the form of ���� � �. ���. Here, ���

is a half-sine wave with the period of 4.93 ��� and � � �0 0 0 � 75 100 0 0 0�� (kN). This

load is illustrated in Figure 3. All required data has been provided so far to do modal analysis as

follows:

function [disp,EF]=NM(8,2,0.005,F,ft,M,K,0,omega2,phi)

Displacement and elastic force time history of some stories are shown in Figures 4 and 5. Note

that displacement and elastic force response will be in m and N, respectively. However, I have

shown them in mm and kN.

Figure 3- Decomposing of Loads into Two Components (dt for f(t) is 0.005 sec)

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Figure 4- Displacement Time history of Some Stories considering Two Modes

Figure 5- Elastic Force Time history of Some Stories considering Two Modes

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Example 2: A MDOF Frame Structure under Elcentro Earthquake

A frame structure is shown in Figure 6. It is desired to find lateral displacement and elastic force

time history of node 3 (in X direction) under Elcentro earthquake N-S component (ground

motion record is available online: http://www.vibrationdata.com/elcentro.htm) considering first 3

modes with 5% damping ratio.

Section area, second moment of inertia and elastic modulus of all elements are same which are

650 mm2, 0.17�10

6 mm

4 and 200�10

3 MPa, respectively. Element self-weight and rotational

mass of nodes are neglected. Lumped mass matrix was derived by applied load in the dimension

of 8�8 (gravity and lateral direction) (Table 1). Global stiffness matrix will be 12�12 (12-DOF)

for this 2D frame. By means of static condensation, we can properly eliminate mass-less DOFs

in stiffness matrix which is summarized in Table 2.

Figure 6- Frame Shape and Loading

Table 1- Mass Matrix 8�8 (Unit: kg) Node 2 Node 3 Node 4 Node 5

Ux(r1) Uy(r2) Ux(r3) Uy(r4) Ux(r5) Uy(r6) Ux(r7) Uy(r8)

122.3242 0 0 0 0 0 0 0

0 122.3242 0 0 0 0 0 0

0 0 122.324159 0 0 0 0 0

0 0 0 122.3242 0 0 0 0

0 0 0 0 122.3242 0 0 0

0 0 0 0 0 122.3242 0 0

0 0 0 0 0 0 122.3242 0

0 0 0 0 0 0 0 122.3242

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Table 2- Condensed Stiffness Matrix without P-∆ Effect 8�8 (Unit: N, mm) Node 2 Node 3 Node 4 Node 5

Ux(r1) Uy(r2) Ux(r3) Uy(r4) Ux(r5) Uy(r6) Ux(r7) Uy(r8)

15333.84 15317.5 -15323.8 -15314.9 -0.32002 -2.90886 0.556449 0.32002

15317.5 58655.22 -15319.4 -15323.4 0.184047 1.672916 -0.32002 -0.18405

-15323.8 -15319.4 58655.22 15317.92 -43333.1 1.672916 -0.32002 -0.18405

-15314.9 -15323.4 15317.92 15329.35 -1.67292 -7.65064 2.90886 1.672916

-0.32002 0.184047 -43333.1 -1.67292 58655.22 -15317.9 -15323.8 15319.41

-2.90886 1.672916 1.672916 -7.65064 -15317.9 15329.35 15314.92 -15323.4

0.556449 -0.32002 -0.32002 2.90886 -15323.8 15314.92 15333.84 -15317.5

0.32002 -0.18405 -0.18405 1.672916 15319.41 -15323.4 -15317.5 58655.22

Note: Change the UNIT to N/m

Table 3- Mode Shape (Φ)

Mode1 Mode2 Mode3 Mode4 Mode5 Mode6 Mode7 Mode8

0.723607 -30.639 -3.42432 4.09675 -1.00044 0.484086 -1.00004 -0.25215

4.35E-05 -0.00632 -0.00081 -2.6451 1.366837 1.877068 -2.19516 -0.40269

1 1 1 1 1 1 1 1

-0.27623 -31.6572 -4.4256 -4.09384 0.999798 -0.48417 1.000152 0.252098

1 0.998295 1.000088 -0.99981 1.000012 -1 1.000012 -1

0.334075 -32.4779 4.254245 -4.09407 -0.99977 -0.48415 -1.00015 0.252112

0.665804 33.45968 -3.25315 -4.09688 -1.00031 -0.4841 -1.00001 0.252139

-3.02E-05 -0.00607 0.000737 -2.6453 -1.36677 1.877095 2.195148 -0.40269

Note: Phi is normalized to be unity in the lateral direction of node 3

� �

005+9.4734e 0 0 0 0 0 0 0

005+6.5069e 0 0 0 0 0 0

005+4.7739e 0 0 0 0 0

005+2.046e 0 0 0 0

005+1.3905e 0 0 0

204.44 0 0

79.014 0

19.586

(rad/s)2

Dynamic properties are shown above. First 4 mode shapes are also plotted in Figure 7. We

will use only first 3 mode shapes and natural frequencies. Note that since stability condition

for numerical solution will not meet in modes above three, we have to consider only three

modes (higher modes have very small periods in this problem).

Sym.

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Figure 7- Mode Shape of first 4 modes

In the case of earthquake, � � ��. . �1 1 1… �� where g=9.81m/s2. Also, ��� � "#���

where "#��� is ground acceleration normalized to g. ��� is plotted in Figure 8.

� � ��. . �1 1 1… �� � �1200�1 1 1 1 1 1 1 1��

Figure 8- Ground Acceleration of N-S Component of Elcentro Earthquake in g (f(t))

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Recall NM Function:

function [disp,EF]=NM(8,3,0.02,F,ft,M,K,5,omega2,phi)

Responses are plotted in Figure 9 for node 3 in X-direction with 5% damping ratio.

Figure 9- Response of Node 3 of Frame Structure in X-Dir. to Elcentro N-S Component

Hints:

plot(elcentro(:,1),disp(:,3)*1000); % Change unit from m to mm plot(elcentro(:,1),EF(:,3));

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Verification:

In this section accuracy of this function will be investigated by comparing the results with the

example 13.2.6 detailed on page 485 of “Dynamic of Structures, Theory and Application to

Earthquake Engineering” (Chopra, 1995) which is a five-story shear frame.

System properties. The lumped mass mj = m = 100 kips/g at each floor (g=386.22 in/s2), the

lateral stiffness of each story is kj = k = 31.54 kips/in., and the height of each story is 12 ft. The

damping ratio for all natural modes is $%= 5%. The mass matrix M, stiffness matrix K, natural

frequencies, and natural modes (normalized to 1 at roof) of this system are as follow:

)/.(

0.2589 0 0 0 0

0 0.2589 0 0 0

0 0 0.2589 0 0

0 0 0 0.2589 0

0 0 0 0 0.2589

2 inskipsM

=

)/(

31.5400 31.5400- 0 0 0

31.5400- 63.0800 31.5400- 0 0

0 31.5400- 63.0800 31.5400- 0

0 0 31.5400- 63.0800 31.5400-

0 0 0 31.5400- 63.0800

inkipsK

=

1.0000 1.0000 1.0000 1.0000 1.0000

2.6825- 1.8308- 0.7154- 0.3097 0.9190

3.5133 0.5211 1.2036- 0.5944- 0.7635

3.2287- 1.3979 0.3728 1.0882- 0.5462

1.9190 1.6825- 1.3097 0.8308- 0.2846

22)/(

448.6144 0 0 0 0

0 344.8605 0 0 0

0 0 208.9717 0 0

0 0 0 84.0919 0

0 0 0 0 9.8694

srad

For the given K and M, the natural periods are Tn = 2.0, 0.6852, 0.4346, 0.3383, and 0.2966 sec.

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Chopra used all 5 modes in modal analysis. Excitation is Elcentro earthquake detailed in

example 2 of this script. Therefore, ��� � "#��� and

� � ��. . �1 1 1… �� � �99.9924 � �1 1 1 1 1 ��

Desired data are available now and then we can recall NM

function [disp,EF]=NM(5,5,0.02,F,ft,M,K,5,omega2,phi)

Figure 10 illustrates a comparison of the roof displacement, roof elastic force and base shear with

this function and Chopra’s solution set. Minor differences may refer to the different types of

analytical algorithms as well as accuracy in digits in current MATLAB version (R2009a)

comparing to Chopra’s software about a decade ago.

a) b)

Figure 10- Comparison of Responses between (a) MN Function and (b) Chopra’s Solution (Chopra 1995, pp

485)

Reference

1. Humar J. L., “Dynamic of Structures”, Prentice Hall, 1990

2. Chopra A. K., “Dynamic of Structures, Theory and Application to Earthquake Engineering”, Prentice

Hall, 1995

3. MATLAB, The MathWorks Inc., 2009