Holooly Plus Logo

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
التقاط
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

(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

k_1=\frac{12 E I}{L^3}=\frac{12 \times 200 \times 10^9 \times 9874.6 \times 10^4}{5^3 \times 10^{12}}=1895923  N / m

Right column base pinned

k_2=\frac{3 E I}{L^3}=473981  N / m

Hence total stiffness = 2369904 N/m
(c) Dynamic characteristics of the structure

\begin{aligned}& \omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{2369904}{26065}}=9.53  rad / s \\& T=\frac{2 \pi}{\omega_n}=\frac{2 \pi}{9.52}=0.659  s\end{aligned}

(d) Time step

\Delta t \angle \Delta t_c r=\frac{T}{\pi}=\frac{0.659}{\pi}=0.209

or

\Delta t=\frac{T}{10}=\frac{0.659}{\pi}=0.659  s

Use 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
التقاط
التقاط
The 'Blue Check Mark' means that either the MATLAB code/script/answer provided in the answer section has been tested by our team of experts; or the answer in general has be fact checked.

Learn more on how do we answer questions.

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’)

Related Answered Questions