Question 7.2: A single story shear frame shown in Fig. 7.4a is subjected t......
A single story shear frame shown in Fig. 7.4a is subjected to arbitrary excitation force specified in Fig. 7.4b. The rigid girder supports a load of 25.57 kN/m.
Assume the columns bend about their major axis and neglect their mass, and assuming damping factor of ρ = 0.02 for steel structures, E = 200 GPa. Write a computer program for the central difference method to evaluate dynamic response for the frame. Plot displacement u(t), velocity v(t) and acceleration a(t) in the interval 0 ≤ t ≤ 5 s.
Table 7.3 Central difference method
| \ddot{u}_0=\frac{F_0-c \dot{u}_0-k u_0}{m} | Step 1 |
| u_{-1}=u_0-\Delta t\left(\dot{u}_0\right)+\frac{\left(\Delta t^2\right)}{2} \ddot{u}_0 | Step 2 |
| \hat{k}=\frac{m}{\Delta t^2}+\frac{c}{2 \Delta t} | Step 3 |
| a=\frac{m}{\Delta t^2}-\frac{c}{2 \Delta t} | Step 4 |
| b=k-\frac{2 m}{(\Delta t)^2} | Step 5 |
| Calculation of time step i
\hat{F}_i=F_i-a u_{i-1}-b u_i |
Step 6 |
| u_{i+1}=\hat{F}_i / \hat{k} | Step 7 |
| Calculate \dot{u}_i=\frac{u_{i+i}-u_{i-1}}{2 \Delta t}
\ddot{u}_i=\frac{u_{i+i}-2 u_i-u_{i-1}}{\Delta t^2} |
Step 8 |
(a) The total load on the beam = 25.57 × 10 = 255.7 kN
\text { Mass }=m=\frac{255.7 \times 10^3}{9.81}=26065 kg(b) Stiffness of the frame (shear frame)
Left column base fixed
Right column base pinned
k_2=\frac{3 E I}{L^3}=473981 N / mHence total stiffness = 2369904 N/m
(c) Dynamic characteristics of the structure
(d) Time step
\Delta t \angle \Delta t_c r=\frac{T}{\pi}=\frac{0.659}{\pi}=0.209or
\Delta t=\frac{T}{10}=\frac{0.659}{\pi}=0.659 sUse time step of 0.05 s
(e) C_c=2 \sqrt{ km }
\begin{aligned}C & =\rho 2 \sqrt{ } km \\& =2 \times 0.02 \sqrt{ } 2369904 \times 26065 \\& =9941.5 N . sec / m\end{aligned}Table 7.4 gives the displacement, velocity and acceleration up to 1 s.
The displacement, velocity and acceleration response are shown in Fig.7.5. The computer program in MATLAB is given below.
Program 7.2: MATLAB program for dynamic response of SDOF using central difference method
Table 7.4 values of u,v and a for Example 7.2
| a | V | U | t | a | V | U | t |
| 0.3594 | -0.0889 | 0.0015 | 0.55 | 0.7673 | 0 | 0 | 0 |
| 0.7886 | -0.0602 | -0.0025 | 0.60 | 0.6664 | 0.0358 | 0.001 | 0.05 |
| 0.8801 | -0.0185 | -0.0045 | 0.65 | 0.4174 | 0.0629 | 0.0036 | 0.10 |
| 0.7718 | 0.0228 | -0.0044 | 0.70 | 0.0791 | 0.0754 | 0.0073 | 0.15 |
| 0.4717 | 0.0544 | -0.0023 | 0.75 | -0.2700 | 0.0704 | 0.0110 | 0.20 |
| 0.1062 | 0.0693 | 0.0011 | 0.80 | -0.5528 | 0.0500 | 0.0143 | 0.25 |
| -0.2960 | 0.0646 | 0.0047 | 0.85 | -0.7052 | 0.0185 | 0.0161 | 0.30 |
| -0.6237 | 0.0416 | 0.0075 | 0.90 | -0.6959 | -0.0165 | 0.0162 | 0.35 |
| -0.8051 | 0.0059 | 0.0088 | 0.95 | -0.6821 | -0.0510 | 0.0145 | 0.40 |
| -0.7256 | -0.0324 | 0.0081 | 0.100 | -0.5750 | -0.0809 | 0.0111 | 0.45 |
| -0.0831 | -0.0831 | 0.0064 | 0.50 |
Script File
% **********************************************************
% DYNAMIC RESPONSE USING CENTRAL DIFFERENCE METHOD
% **********************************************************
ma=26065;
k=2369904.0;
wn=sqrt(k/ma)
r=0.02;
c=2.0*r*sqrt(k*ma)
u(1)=0;
v(1)=0;
tt=5;
n=100;
n1=n+1
dt=tt/n;
td=.9;
jk=td/dt;
%***********************************************************
% LOADING IS DEFINED HERE
%***********************************************************
for m=1:n1
p(m)=0.0;
end
t=-dt
for m=1:8;
t=t+dt;
p(m)=20000;
end
p(9)=16000.0
for m=10:12
t=t+dt
p(m)=12000.0
end
for m=13:19
t=t+dt
p(m)=12000.0*(1-(t-0.6)/.3)
end
an(1)=(p(1)-c*v(1)-k*u(1))/ma
up=u(1)-dt*v(1)+dt*dt*an(1)/2
kh=ma/(dt*dt)+c/(2.0*dt)
a=ma/(dt*dt)-c/(2.0*dt)
b=k-2.0*ma/(dt*dt)
f(1)=p(1)-a*up-b*u(1)
u(2)=f(1)/kh
for m=2:n1
f(m)=p(m)-a*u(m-1)-b*u(m)
u(m+1)=f(m)/kh
end
v(1)=(u(2)-up)/(2.0*dt)
for m=2:n1
v(m)=(u(m+1)-u(m-1))/(2.0*dt)
an(m)=(u(m+1)-2.0*u(m)+u(m-1))/(dt*dt)
end
n1p=n1+1
for m=1:n1p
s(m)=(m-1)*dt
end
for m=1:n1
x(m)=(m-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(x,v,‘k’);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response velocity v in m/sec’)
title(‘ dynamic response’)
figure(3);
plot(x,an,‘k’);
xlabel(‘ time (t) in seconds’)
ylabel(‘ Response acceleration a in m/sq.sec’)
title(‘ dynamic response’)