Question 7.3: A steel water tank shown in Fig. 7.6a is analysed as an SDOF......

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

A steel water tank shown in Fig. 7.6a is analysed as an SDOF system having a mass on top of cantilever which acts as a spring and dashpot for damping.

A blast load of P(t) is applied as shown in Fig. 7.6b. The values of the force are given in Table 7.6. Draw displacement, velocity and acceleration responses up to 0.5 s. The damping for steel may be assumed to be 2% of critical damping.

Table 7.6 Blast load at various times

0.10	0.09	0.08	0.07	0.06	0.05	0.04	0.03	0.02	0.01	0	t
0	26.9	53.4	89	142	213	284	364	445	267	0	$10^3 \times P(t)$

Step-by-Step

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Given
W = 133.5 kN

\begin{aligned}& \text { Mass }=m=\frac{133500}{9.81}=13608.5  kg \\& k=17500 \times 10^3  N / m \\& \omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{17500 \times 10^3}{13608.5}}=35.8  rad / s \\& 35.8 / 2 \pi=5.7 cycles / s\end{aligned}

Fundamental period T = 1/5.7 = 0.175 s.

\Delta t_{c r}=\frac{T}{\pi}=\frac{0.175}{\pi}=0.055 s

Δt = T/10 = 0.0175s.
Use Δt = 0.01s

Figure 7.7 shows the displacement, velocity and acceleration response for the tank. The values are given in Table 7.7 and the program is given below.

Program 7.3: MATLAB program for dynamic response of SDOF by
Runge–Kutta method

Table 7.7 Displacement, velocity and acceleration at various times for Example 7.3

a	v	u	t	a	v	u	t
-9.2278	-0.9789	0.0083	0.11	0	0	0	0
3.6697	-1.0071	-0.0011	0.12	18.5579	0.1605	0.0007	0.01
15.9694	-0.9077	-0.0195	0.13	27.4821	0.4375	0.0036	0.02
26.0770	-0.6991	-0.0250	0.14	14.2647	0.6282	0.0090	0.03
32.7473	-0.3973	-0.0273	0.15	-0.2441	0.6789	0.0157	0.04
25.1872	-0.0544	-0.0261	0.16	-13.6410	0.5909	0.0221	0.05
33.1480	0.2909	-0.0216	0.17	-24.9000	0.3783	0.0271	0.06
24.9539	0.5946	-0.0145	0.18	-31.3880	0.0805	0.0294	0.07
17.442	0.8188	-0.0056	0.19	-32.4410	-0.2509	0.0286	0.08
5.8649	0.9365	0.0038	0.20	-28.6582	-0.5663	0.0244	0.09
				-21.2242	-0.8250	0.0174	0.10

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Script File

%********************************************************************
% DYNAMIC RESPONSE DUE TO EXTERNAL LOADING RUNGE
KUTTA METHOD
%********************************************************************
ma=13608.5;
k=17500000;
wn=sqrt(k/ma)
r=0.02;
c=2.0*r*sqrt(k*ma)
u(1)=0;
v(1)=0;
tt=.5;
n=50;
n1=n+1
dt=tt/n;
td=.1;

jk=td/dt;
%**************************************************************
% EXTERNAL LOADING IS DEFINED HERE
%**************************************************************
for m=1:n1

p(m)=0.0;
end
p(2)=267000.0
p(3)=445000.0
p(4)=364000.0
p(5)=284000.0
p(6)=213000.0
p(7)=142000.0
p(8)=89000.0

p(9)=53400.0
p(10)=26700.0
an(1)=(p(1)-c*v(1)-k*u(1))/ma
t=0.0
for i=2:n1
ui=u(i-1)
vi=v(i-1)
ai=an(i-1)
d(1)=vi
q(1)=ai
for j=2:3
l=0.5
x=ui+l*dt*d(j-1)
d(j)=vi+l*dt*q(j-1)
q(j)=(p(i)-c*d(j)-k*x)/ma
end
j=4
l=1
x=ui+l*dt*d(j-1)
d(j)=vi+l*dt*q(j-1)
q(j)=(p(i)-c*d(j)-k*x)/ma
u(i)=u(i-1)+dt*(d(1)+2.0*d(2)+2.0*d(3)+d(4))/6.0
v(i)=v(i-1)+dt*(q(1)+2.0*q(2)+2.0*q(3)+q(4))/6.0
an(i)=(p(i)-c*v(i)-k*u(i))/ma
end
for i=1:n1
s(i)=(i-1)*dt
end
figure(1);
plot(s,u,‘k’);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response displacement u in m’)
title(‘ dynamic response’)
figure(2);
plot(s,v,‘k’);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response velocity v in m/sec’)
title(‘ dynamic response’)
figure(3);
plot(s,an,‘k’);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response acceleration a in m/sec’)
title(‘ dynamic response’)