Abstract: In this chapter, earthquake excitation problems are treated as base excitation problems. In many engineering applications, one requires maximum absolute quantities experienced by the structure during the earthquake of interest. In that respect, the response spectrum method is ideally suited to designing structures against earthquake motion. It is shown how the tripartite plot is useful for reading all the spectral quantities for a given period. The construction of a Newmark–Hall design spectrum is illustrated and the distinction is made between design and response spectra. In all the codes, site-specific response spectra are recommended. Finally inelastic design spectra are also discussed.

**Earthquake
response spectra**

**Abstract: **In this chapter, earthquake
excitation problems are treated as base** **excitation problems. In many
engineering applications, one requires maximum absolute quantities experienced
by the structure during the earthquake of interest. In that respect, the
response spectrum method is ideally suited to designing structures against earthquake
motion. It is shown how the tripartite plot is useful for reading all the
spectral quantities for a given period. The construction of a Newmark–Hall
design spectrum is illustrated and the distinction is made between design and
response spectra. In all the codes, site-specific response spectra are
recommended. Finally inelastic design spectra are also discussed.

**Key words: **accelerograph, response spectrum,
spectral quantity, pseudo** **spectral quantity, ductility, time history.

**Introduction **

The earthquake response problem is essentially a base
excitation problem similar to the one discussed in the earlier chapter for
single-degree-of-freedom (SDOF) systems. The equation of motion is written as

Thus for any arbitrary acceleration of the supports the
relative displacement of the mass can be computed using the *Duhamel integral*
as

From Eq. 17.2 it is seen that the
relative response of the structure is characterized by its natural frequency,
damping factor and the nature of base excitation.

Generally we use undamped natural frequency instead of damped
frequency and the negative sign is ignored (the sense of response has no
significance in earthquake analysis)

*v*(*t*) =* R*(*t*)
is called the* earthquake response integral*. The relative displacement*
*is important since it is required to calculate base shear.

It is noticed that base shear represented in Eq. 17.5 is
equivalent to restoring force *F _{s}*(

The absolute total acceleration of the mass is obtained by
adding relative acceleration with ground acceleration as

The total acceleration has many important applications. It can
most easily be measured experimentally during a strong earthquake. When an *accelerograph*
is located in a structure, it records in close approximation the total
acceleration at that point. From this we can calculate inertia force

Equations 17.2, 17.6 and 17.7 represent the earthquake time
history response for an SDOF system. Once we know base shear *V* (*effective
earthquake* *force*), we can design a system.

**Earthquake response spectra **

It has been seen from earlier chapters that the evaluation of *dynamic
response* (displacement, velocity and acceleration) at every instant of time
during an earthquake requires significant computational effort even for
relatively simple structural systems. However, for many engineering
applications we require maximum absolute quantities experienced by the structure
during the earthquake. These are commonly referred to as spectral displacement *S _{d}*,
spectral velocity

are the maximum absolute values of relative displacement, relative velocity and
total acceleration determined from Eq. 17.2, 17.6 and 17.7. However, these
quantities are generally determined from numerical integration techniques.
Plots of Sd, Sv, Sa versus undamped natural period of vibration for various
damping ratios are called earthquake response spectra. We can construct an
earthquake response spectrum, say, for Northridge earthquake (see Fig 17.1a) by
considering a series of oscillators (inverted pendulums) as shown in Fig. 17.1b
with varying periods of vibration attached with movable base. The base is
subjected to same ground motion as that of Northridge earthquake. The maximum
response for each pendulum (Sd, Sv, Sa) is plotted against the natural period
for a particular value of damping. Such curves are the response spectra and are
very useful for design. Figure 17.2 shows response spectra for relative
displacement, relative velocity and total acceleration for the Northridge
earthquake. These quantities are
generally determined by numerical evaluation using any of the numerical
techniques developed in Previous Pages. Plots of Sd, Sv and *S _{a} *versus
undamped natural period of vibration or natural frequency for

**Program 17.1: MATLAB program
for drawing spectra for any specified earthquake **

%********************************************************** %
WILSON’S RECURRENCE FORMULA TO DRAW SPECTRA FOR ANY EARTHQUAKE MOTION %**********************************************************

clc;close
all;

%give initial displacement and initial velocity u(1)=0;

v(1)=0;

%tt=50 sec n=2500 for nridge %tt=30 n=1500 for elcentro ns
tt=50.0;

n=2500;

n1=n+1;

dt=tt/n;

mass=1;

%**********************************************************

% EARTHQUAKE
DATA FILE FOR NORTH-RIDGE EQ IS READ FROM EXCEL DATA FILE

% DATA FILE
CONSISTS OF TIME AND THE CORRESPONDING A/G VALUES

%DATA FILE
CHOPRA CONSISTS OF ELCENTRO NS DATA

EQDATA CONTAINS NORTHRIDGE DATA

%**********************************************************

de=xlsread(‘eqdata’);

%de=xlsread(‘chopra’) for i=1:n1;
ug(i)=de(i,2); ug(i)=-ug(i)*9.81; p(i)=ug(i)*mass;

end;

%define damping ratios for which response
is required rho=[0 0.02 0.05 .1 0.2];

for jj=1:5 r=rho(jj);

mass=1; for ii=1:200

tn=0.05*ii;

wn=2.0*pi/tn;

ss(ii)=tn;

k=mass*wn^2;

c=2.0*r*sqrt(k*mass);
wd=wn*sqrt(1-r^2);

a=exp(-r*wn*dt)*(r*sin(wd*dt)/sqrt(1-r^2)+cos(wd*dt));
b=exp(-r*wn*dt)*(sin(wd*dt))/wd;
c2=((1-2*r^2)/(wd*dt)-r/sqrt(1-r^2))*sin(wd*dt)-(1+2*r/(wn*dt))*cos(wd*dt);
c=(1/k)*(2*r/(wn*dt)+exp(-r*wn*dt)*(c2));

d 2 = e x p ( - r * w n * d t ) *
( ( 2 . 0 * r ^ 2 - 1 ) / ( w d * d t ) * s i n ( w d * d t ) + 2 . 0 * r /
(wn*dt)*cos(wd*dt));

d=(1/k)*(1-2.0*r/(wn*dt)+d2);
ad=-exp(-r*wn*dt)*wn*sin(wd*dt)/(sqrt(1-r^2));
bd=exp(-r*wn*dt)*(cos(wd*dt)-r*sin(wd*dt)/sqrt(1-r^2));

c 1 = e x p ( - r * w n * d t ) *
( ( w n / s q r t ( 1 - r ^ 2 ) + r / ( d t * s q r t ( 1 - r^2)))*sin(wd*dt)+cos(wd*dt)/dt);

cd=(1/k)*(-1/dt+c1);
d1=exp(-r*wn*dt)*(r*sin(wd*dt)/sqrt(1-r^2)+cos(wd*dt)); dd=(1/(k*dt))*(1-d1);

for m=2:n1 u(m)=a*u(m-1)+b*v(m-1)+c*p(m-1)+d*p(m);
v(m)=ad*u(m-1)+bd*v(m-1)+cd*p(m-1)+dd*p(m); ta(m)=(-c*v(m)-k*u(m))/mass;

end dmf(ii,jj)=max(abs(u));
vmf(ii,jj)=max(abs(v)); amf(ii,jj)=max(abs(ta)); end

end

for m=1:n1 s(m)=(m-1)*dt;

end figure(1);

plot(s,ug/9.81,‘K’);

xlabel(‘ time (t) in seconds’);
ylabel(‘ ground acceleration/g’); title(‘ North Ridge EQ data’); figure(2);

for ii=1:5 plot(ss,dmf(:,ii),‘K’); hold on

end

xlabel(‘
period T in seconds’);

ylabel(‘
Spectral displacement Sd in m’);

title(‘
Response spectrum for Sd North Ridge EQ)’);

legend(‘ dam=0’,‘dam=0.02’,‘dam=0.05’, ‘dam=0.1’,‘dam=0.2’) figure(3);

for ii=1:5 plot(ss,vmf(:,ii),‘K’); hold on

end

xlabel(‘ period T in seconds’);

ylabel(‘
Spectral velocity Sv in m/sec’);

title(‘
Response spectrum for Sv (North Ridge EQ)’);

legend(‘ dam=0’,‘dam=0.02’,‘dam=0.05’, ‘dam=0.1’,‘dam=0.2’)
figure(4);

for ii=1:5 plot(ss,amf(:,ii)/9.81,‘K’); hold on

end

xlabel(‘
period T in seconds’);

ylabel(‘
Spectral total acceleration Sa/g’);

title(‘ Response spectrum for Sa (total) (North Ridge EQ)’);
legend(‘ dam=0’,‘dam=0.02’,‘dam=0.05’, ‘dam=0.1’,‘dam=0.2’)
fid=fopen(‘sv.out’,‘w’)

for
jj=1:200

fprintf(fid,‘
%6.3f %6.2f %6.2f %6.2f %6.2f %6.2f\n’...

,ss(jj),vmf(jj,1),vmf(jj,2),vmf(jj,3),vmf(jj,4),vmf(jj,5));
end

fclose(fid)

Example 17.1

An industrial building is shown
in Fig. 17.3. Idealize the structure as an SDOF system. Assume the structure
acts as a braced frame in the EW direction (having a total of six braced bays)
and *unbraced shear frame* **(**with column base) pinned in the NS
direction. Assume all columns bend about the major axis to the NS direction.

The vertical cross-bracings in
the EW direction are 25.4 mm diameter steel rods. The dead weight of the
structure is 1291.95 kN which is concentrated at the base of the roof trusses.
Moment of inertia of columns 8.6992 × 10^{7}
mm^{4}. The height of the building may be assumed as 4.2672 m and the
damping is 5% of critical damping. Consider Northridge earthquake.

(a) Determine
the natural period for the NS and EW directions.

(b) Conduct a
time history analysis of the structure in both directions. Use the NS component
of the 17 January, 1994 Northridge earthquake shown in Fig. 17.1a as input.

Solution

(a) *Natural
period*. Mass of the structure

(b) *Time history response*.
The parameters in the time history response are relative displacement, *u*(*t*),
relative velocity is *u***˙**(*t* ), absolute total acceleration
is *u***˙˙*** _{t}* (

The equation of moment in NS direction

Equation
of moment in EW direction

*u***˙˙*** *+* *2* *×* *0.05* *×* *20* u***˙*** *+* *20* *^{2}* u *=* *−*u***˙˙**_{g}* *(*t *)* *

i.e. *u***˙˙** + 2 *u***˙**
+ 400 *u* = −*u***˙˙*** _{g}* (

The value can be obtained for the
Northridge earthquake by integrating above two expressions.

As an example. The response displacement, velocity and total
acceleration are shown in Fig. 17.4 and Table 17.1 gives maximum response in NS
direction.

Example 17.2

Construct response spectra for
the NS component of E1 Centro earthquake (see Fig. 17.5a). Consider damping
factors 0, 0.02, 0.05, 0.1 and 0.2.

Solution

The spectral displacement *S _{d}*, spectral
velocity

for *ρ* = 0.02, 0.05 and 0.1 from which
maximum responses are determined. The response spectral *S _{d}*,

Let us define pseudo-spectral velocity and pseudo spectral
acceleration as *S _{pv}*,

Hence the spectral relationship significantly expedites the
construction of earthquake response spectra. Evaluation of spectra displacement
*s _{d}* after numerical integration to obtain time history
response, the corresponding pseudo spectral velocity

Then for a given frequency or for
a given period all the spectral quantities can be read simultaneously for the
same tripartite plot. A *tripartite plot* of *s _{d}*,

*s _{pvr}*,

**Program
17.2: MATLAB program to draw tripartite plot **

%**********************************************************

% to draw
elastic design spectra from the earthquake data

% read the
response spectrum values in this program

%**********************************************************

for
k=.00001:.00001:.0001

x=0.01:1:100

t=log(2*pi*k)-log(x)

y=exp(t)

loglog(x,y,‘k’),grid
on

hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’)

hold on

end

for
k=.0001:.0001:.001

x=0.01:1:100

t=log(2*pi*k)-log(x)

y=exp(t)

loglog(x,y,‘k’),grid
on

hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) hold on

end

for k=.001:.001:.01 x=0.01:1:100
t=log(2*pi*k)-log(x) y=exp(t) loglog(x,y,‘k’),grid on hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) hold on

end

xlabel(‘ period in secs’)

ylabel(‘ spectral velocity sv in
cm/sec’) for k=.01:.01:.1

x=0.01:1:100 t=log(2*pi*k)-log(x)
y=exp(t) loglog(x,y,‘k’),grid on hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) hold on

end

for k=.1:.1:1 x=0.01:1:100
t=log(2*pi*k)-log(x) y=exp(t) loglog(x,y,‘k’),grid on hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) hold on

end

for k=1:1:10 x=0.01:1:100 t=log(2*pi*k)-log(x) y=exp(t)

loglog(x,y,‘k’),grid on hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) hold on

end

for k=10:10:100 x=0.01:1:100
t=log(2*pi*k)-log(x) y=exp(t) loglog(x,y,‘k’),grid on hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) hold on

end

for k=100:100:1000 x=0.01:1:100
t=log(2*pi*k)-log(x) y=exp(t) loglog(x,y,‘k’),grid on hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) hold on

end

for k=1000:1000:10000
x=0.01:1:100 t=log(2*pi*k)-log(x) y=exp(t) loglog(x,y,‘k’),grid on hold on

t=log(k*9.81/(2*pi))+log(x)

y=exp(t)

loglog(x,y,‘k’) end

axis([0.01 100 0.02 500]) %
d=xlsread(‘svdata’); sv=‘sv.out’

d=load(sv)

plot(d(:,1),100*d(:,2),‘k’)

plot(d(:,1),100*d(:,3),‘k’)

plot(d(:,1),100*d(:,4),‘k’)

plot(d(:,1),100*d(:,5),‘k’)

plot(d(:,1),100*d(:,6),‘k’)

text(0.2,0.02,‘0.001’);

text(0.6,0.1,‘0.01’);

text(2,0.3,‘0.1’);

text(7,1,‘1’);

text(20,3,‘10’);

text(80,10,‘100’) text(20,1,‘sd
in cm’) xlabel(‘ period in sec’) ylabel(‘ sv in cm/sec’) text(0.01,200,‘100’)
text(0.01,20,‘10’) text(0.01,2,‘1’) text(0.02,0.4,‘0.1’) text(0.07,0.1,‘0.01’)
text(.02,0.8,’sa/g’) gtext(‘ no damping’) gtext(‘ damping=2%’) gtext(‘
damping=5%’) gtext(‘ damping=10%’) gtext(‘ damping=20%’)

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