Question 7.4: The shear frame shown in Fig. 7.9a is subjected to the expon......

Structural Dynamics of Earthquake Engineering: Theory and Application Using Mathematica and Matlab [1014074]

The shear frame shown in Fig. 7.9a is subjected to the exponential pulse force shown in Fig 7.9b. Write a computer program for the average acceleration method (incremental formulation) to evaluate the dynamic response of the frame. Plot the time histories for displacement u(t), velocity $\dot{u}(t)$ and acceleration $\ddot{u}(t)$ in the time interval 0-3 s. Assume E = 200 GPa, W =1079.1 kN, ρ = 0.07, $F_0=450 kN \text { and } t_{ d }=0.75 s$ and use a time step Δt of 0.01 s.

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I for ISWB 600 @1337 = 106198.5e $^4mm^4$

Given

\begin{aligned}& \text { Mass }=m=110000  kg \\& k=\frac{3 \times 200 \times 10^9 \times 106198.5 \times 10^4}{10^{12} \times 5^3} \\& +\frac{12 \times 200 \times 10^9 \times 106198.5 \times 10^4}{10^{12} \times 8^3} \\&\end{aligned}

= 5097528 + 4978054
= 10075582 N/m

\begin{aligned}& \omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{10075582}{110 \times 1000}}=9.57  rad / s \\& T=\frac{2 \pi}{\omega_n}=\frac{2 \pi}{9.57}=0.65  s \\& \Delta t \leq \frac{T}{10} \\& \Delta t \leq 0.065  s\end{aligned}

We use Δt = 0.01 s.
The displacement velocity and acceleration response are shown in Fig.7.10 and the values are given in Table 7.9. The program using MATLAB for constant acceleration method is given below.
Program 7.4: MATLAB program for dynamic response by constant acceleration method

Table 7.9 Displacement, velocity and acceleration response for Example 7.4

a	v	u	t	a	v	u	t
0.5505	0.2701	0.0185	0.11	4.0909	0	0	0
0.1883	0.2738	0.0212	0.12	3.8587	0.0397	0.0002	0.01
-0.1669	0.2739	0.0239	0.13	3.6001	0.0770	0.0008	0.02
-0.5122	0.2705	0.0266	0.14	3.3177	0.1116	0.0017	0.03
-0.8446	0.2637	0.0293	0.15	3.0141	0.1433	0.0030	0.04
-1.1615	0.2537	0.0319	0.16	2.6922	0.17178	0.0046	0.05
-1.4602	0.2406	0.0347	0.17	2.3550	0.1937	0.0064	0.06
-1.7385	0.2246	0.0367	0.18	2.0057	0.2189	0.0085	0.07
-1.9941	0.2059	0.0389	0.19	1.6473	0.2371	0.0108	0.08
-2.2253	0.18580	0.0408	0.20	1.2832	0.2518	0.0132	0.09
				0.9165	0.2628	0.0158	0.10

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Script File

% Response by constant acceleration method.
ma=110000;

k=10075582;
wn=sqrt(k/ma)
r=0.07;
c=2.0*r*sqrt(k*ma)
u(1)=0;
v(1)=0;

tt=3;
n=300;
n1=n+1
dt=tt/n;
td=.75;
jk=td/dt;
for m=1:n1
p(m)=0.0;
end
jk1=jk+1
for n=1:jk1
t=(n-1)*dt
p(n)=450000*(1-t/td)*exp(-2.0*t/td)
end
an(1)=(p(1)-c*v(1)-k*u(1))/ma
kh=k+4.0*ma/(dt*dt)+2.0*c/dt
for i=1:n1
s(i)=(i-1)*dt
end
for i=2:n1
ww=p(i)-p(i-1)+(4.0*ma/dt+2.0*c)*v(i-1)+2.0*ma*an(i-1)
xx=ww/kh
yy=(2/dt)*xx-2.0*v(i-1)
zz=(4.0/(dt*dt))*(xx-dt*v(i-1))-2.0*an(i-1)
u(i)=u(i-1)+xx
v(i)=v(i-1)+yy
an(i)=an(i-1)+zz
end
figure(1);

plot(s,u);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response displacement u in m’)
title(‘ dynamic response’)
figure(2);
plot(s,v);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response velocity v in m/sec’)
title(‘ dynamic response’)
figure(3);
plot(s,an);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response acceleration a in m/sec’)
title(‘ dynamic response’)
figure(4);
plot(s,p)
xlabel(‘ time (t) in seconds’)
ylabel(‘ force in Newtons’)
title(‘ Excitation Force’)