MATLAB Examples

# Chopra (2012): DRSA for shear buiding subjected to the El Centro earthquake motion

## Statement of the problem

• Chopra (2012), Section 12.8: Consider the structure of Fig. 12.8.1: a uniform five-story shear building (i.e., flexurally rigid floor beams and slabs) with lumped mass m at each floor, and same story stiffness k for all stories.
• Chopra (2012), Section 13.2.6: The structure is subjected to the El Centro ground motion (Chopra (2012), Fig. 6.1.4). The lumped mass at each floor, the lateral stiffness of each story is , and the height of each story is 12 ft. The damping ratio for all natural modes is .

## Initialization of structural input data

Set the storey height of the structure in ft.

h=12; 

Set the number of eigenmodes of the structure, which is equal to the number of its storeys.

neig=5; 

Set the lateral stiffness of each storey in kips/inch.

k=31.54; 

Set the lumped mass at each floor (g=386.4 inch/sec^2).

m=100/9.81*0.0254; 

## Calculation of structural properties

Calculate the stiffness matrix of the structure in kips/inch.

K=k*(diag([2*ones(neig-1,1);1])+diag(-ones(neig-1,1),1)+diag(-ones(neig-1,1),-1)); 

Calculate the mass matrix of the structure.

M=m*eye(neig); 

Set the spatial distribution of the effective earthquake forces. Earthquake forces are applied at all dofs of the structure.

r=ones(5,1); 

## Calculation of response spectrum ordinates

Open file elcentro.dat.

fid=fopen('elcentro.dat','r'); 

Read the text contained in the file elcentro.dat.

text=textscan(fid,'%f %f'); 

Close file elcentro.dat.

fclose(fid); 

Set the time step of the input acceleration time history.

time=text{1,1}; dt=time(2)-time(1); 

Set the input acceleration time history () in inch/sec^2.

xgtt=9.81/0.0254*text{1,2}; 

Set the critical damping ratio of the response spectra to be calculated ()

ksi=0.05; 

Dynamic Response Spectrum Analysis

[U,~,~,f,omega,Eigvec] = DRSA(K,M,r,dt,xgtt,ksi); 

Plot the natural modes of vibration of the uniform five-story shear building

FigHandle=figure('Name','Natural Modes','NumberTitle','off'); set(FigHandle,'Position',[50, 50, 1000, 500]); for i=1:neig subplot(1,neig,i) plot([0;Eigvec(:,i)],(0:h:h*neig)','LineWidth',2.,'Marker','.',... 'MarkerSize',20,'Color',[0 0 1],'markeredgecolor','k') grid on xlabel('Displacement','FontSize',13); ylabel('Height','FontSize',13); title(['Mode ',num2str(i)],'FontSize',13) end 

Compare the eigenmodes with those shown in Figure 12.8.2. in Chopra (2012).

Plot the peak modal displacement response.

FigHandle=figure('Name','Displacements','NumberTitle','off'); set(FigHandle, 'Position', [50, 50, 1000, 500]); for i=1:neig subplot(1,neig,i) plot([0;U(:,i)],(0:h:h*neig)','LineWidth',2.,'Marker','.',... 'MarkerSize',20,'Color',[0 1 0],'markeredgecolor','k') xlim([-max(abs(U(:,i))) max(abs(U(:,i)))]) grid on xlabel('Displacement','FontSize',13); ylabel('Height','FontSize',13); title(['Mode ',num2str(i)],'FontSize',13) end 

Plot the peak modal equivalent static force response.

FigHandle=figure('Name','Equivalent static forces','NumberTitle','off'); set(FigHandle, 'Position', [50, 50, 1000, 500]); for i=1:neig subplot(1,neig,i) plot([0;f(:,i)],(0:h:h*neig)','LineWidth',2.,'Marker','.',... 'MarkerSize',20,'Color',[1 0 0],'markeredgecolor','k') xlim([-max(abs(f(:,i))) max(abs(f(:,i)))]) grid on xlabel('Static force','FontSize',13); ylabel('Height','FontSize',13); title(['Mode ',num2str(i)],'FontSize',13) end 

Calculate the peak modal base shear in kips.

Vb=zeros(1,neig); for i=1:neig Vb(i)=sum(f(:,i)); end Vb 
Vb = 60.4052 24.2322 9.8689 2.8754 0.5834 

Calculate the peak modal base overturning moment in kips-ft.

Mb=zeros(1,neig); for i=1:neig Mb(i)=sum(f(:,i).*(h:h:5*h)'); end Mb 
Mb = 1.0e+03 * 2.5467 -0.3500 0.0904 -0.0205 0.0036 

## Verification of figure 13.8.3 of Chopra (Dynamics of Structures, 2012)

Compare the peak modal responses with those shown in figure 13.8.3 of Chopra (2012).

## Modal combination with the ABSolute SUM (ABSSUM) method.

Peak base shear.

VbAbsSum=ABSSUM(Vb); 

Peak top-story shear.

V5AbsSum=ABSSUM(f(5,:)); 

Peak base overturning moment.

MbAbsSum=ABSSUM(Mb); 

Peak top-story displacement.

u5AbsSum=ABSSUM(U(5,:)); 

## Modal combination with the Square Root of Sum of Squares (SRSS) method.

Peak base shear.

VbSRSS=SRSS(Vb); 

Peak top-story shear.

V5SRSS=SRSS(f(5,:)); 

Peak base overturning moment.

MbSRSS=SRSS(Mb); 

Peak top-story displacement.

u5SRSS=SRSS(U(5,:)); 

## Modal combination with the Complete Quadratic Combination (CQC) method.

Peak base shear.

VbCQC=CQC(Vb,omega,ksi); 

Peak top-story shear.

V5CQC=CQC(f(5,:),omega,ksi); 

Peak base overturning moment.

MbCQC=CQC(Mb,omega,ksi); 

Peak top-story displacement.

u5CQC=CQC(U(5,:),omega,ksi); 

## Assemble values of peak response in a table.

Assemble values of peak response in a cell.

C{1,2}='Vb (kips)'; C{1,3}='V5 (kips)'; C{1,4}='Mb (kip-ft)'; C{1,5}='u5 (in)'; C{2,1}='ABSSUM'; C{2,2}=VbAbsSum; C{2,3}=V5AbsSum; C{2,4}=MbAbsSum; C{2,5}=u5AbsSum; C{3,1}='SRSS'; C{3,2}=VbSRSS; C{3,3}=V5SRSS; C{3,4}=MbSRSS; C{3,5}=u5SRSS; C{4,1}='CQC'; C{4,2}=VbCQC; C{4,3}=V5CQC; C{4,4}=MbCQC; C{4,5}=u5CQC; C 
C = [] 'Vb (kips)' 'V5 (kips)' 'Mb (kip-ft)' 'u5 (in)' 'ABSSUM' [ 97.9651] [ 56.2087] [ 3.0113e+03] [ 7.9562] 'SRSS' [ 65.8938] [ 29.8774] [ 2.5723e+03] [ 6.7964] 'CQC' [ 66.3294] [ 29.1493] [ 2.5695e+03] [ 6.7891] 

## Verification of table 13.8.5 of Chopra (Dynamics of Structures, 2012)

Compare the peak responses with those shown in table 13.8.5 of Chopra (2012).

## Copyright

Copyright (c) 13-Sep-2015 by George Papazafeiropoulos