1. CIVE 801
DYNAMICS OF STRUCTURES I
COURSE PROJECT
Cyle Teal
CIVE 801, Winter 14-15
March 16, 2015
2. Table of Contents
INTRODUCTION .....................................................................................................................2
1. FRAME DESIGN .............................................................................................................2
Girder Sizing ......................................................................................................................... 4
Longitudinal Girder weight.................................................................................................. 4
Deck Weight ......................................................................................................................... 4
Overall Weight.......................................................................................................................5
2. NATURAL FREQUENCY AND PERIOD.......................................................................5
Stiffness ..................................................................................................................................5
Mass....................................................................................................................................... 6
Natural Frequency ................................................................................................................ 6
Natural Period....................................................................................................................... 6
3. UNDAMPED FREE VIBRATION .................................................................................. 6
4. VISCOUSLY DAMPED FREE VIBRATION ...................................................................7
Part (i) ................................................................................................................................... 8
Part (ii) .................................................................................................................................. 9
5. STRUCTURE SUBJECTED TO SINUSOIDAL HORIZONTAL FORCE...................... 10
6. LIMITS FOR FREQUENCY OF VIBRATING MACHINE.............................................11
7. PERIODIC RECTANGULAR WAVE............................................................................ 12
8. STRUCTURAL RESPONSE TO ARBITRARY EXCITATION ...................................... 15
CONCLUSION ........................................................................................................................ 17
APPENDIX............................................................................................................................... 18
1. System Characteristics ................................................................................................. 18
2. Determine natural frequency and period of structure............................................... 18
3. Subject structure to initial conditions......................................................................... 18
4. Assume damping ratios of 2%, 5%, 15%, 70%, 100% and 150% .................................. 19
5. Subject structure to a sinusoidal horizontal force...................................................... 21
6. Limits for the freqency of vibrating machine for TR = 20% condition......................22
3. 7. Fourier representation of periodic rectangular wave.................................................23
8. Response of structure to arbitrary excitation .............................................................25
PAGE 1
4. INTRODUCTION
CIVE 801 covered dynamics of single-degree-of-freedom (SDF) systems. The course
project, detailed herein, required a realistic frame to be designed and idealized as a SDF
system. This report presents the results of the analyses and computations completed to
fulfill the requirements for the project. The MATLAB code used to complete the project
is provided in the appendix.
1. FRAME DESIGN
The structure chosen for the project, a bridge supported by two abutments and two
bents, is an adaptation of the one in Example 1.3 of the course text1 and shown in Figure 1.
Figure 1. 375-ft-long concrete bridge on four supports.
1
Chopra, Anil K. Dynamics of Structures, Theory and Applications to Earthquake Engineering, 4th
ed., 2012.
Pearson Education, Inc.
PAGE 2
5. The bridge can be represented as a frame (see Figure 2) with each bent providing a
longitudinal stiffness 𝑘𝑘bent.
Figure 2. Frame representation of the bridge.
In the problem statement for Example 1.3, it was given that each bent consisted of three
25-ft-tall columns of circular section with 𝐼𝐼𝑦𝑦′ = 𝐼𝐼𝑧𝑧′ = 13 ft4
. These values were assumed
for the project.
The problem also gave the following specifications:
• Box-girder bridge
• Cross-sectional area of the bridge deck = 123 ft2
• Unit weight of concrete = 150 lb/ft3
Of these three, only the last was assumed. A multi-girder design was selected instead of a
box-girder. The multi-girder cross section shown in Figure 3 was assumed for the project.
Four longitudinal plate girders of equal depth are arranged at uniform spacing across the
width of the bridge. The deck slab spans transversely across the longitudinal girders and
cantilevers outside the outer girders. The girders were assumed to be spaced 12 ft on
center and the deck was assumed to have a constant thickness of 1 ft across the 43-ft-
width. Before the weight per unit length of the cross section could be calculated, the
girders had to be realistically sized for a bridge of this span.
Figure 3. Cross section of a multi-girder bridge2
2
Taken from steelconstruction.info
PAGE 3
6. GIRDER SIZING
The longitudinal girders were proportioned using the recommended girder proportions
found on steelconstruction.info3. The following table presents the values used to calculate
the cross-sectional area of each longitudinal girder.
Table 1. Girder Element Sizes.
Element Value (in)
Girder Depth 50
Top flange width 14
Top flange thickness 1
Bottom flange width 25
Bottom flange thickness 2.25
Web 0.75
The sizes in Table 1 resulted in the following girder cross-sectional area
𝐴𝐴st = 105.3 in2
LONGITUDINAL GIRDER WEIGHT
Assuming a density of steel, 𝜌𝜌st = 0.28
lb
in3, the longitudinal girder contribution to the
overall weight per unit length was calculated
𝑤𝑤g = 4 𝜌𝜌st 𝐴𝐴st = 4 �0.28
lb
in3
� (105.3 in2) �
1 kip
1000 lb
� �
12 in
1 ft
� = 1.42
kip
ft
DECK WEIGHT
Assuming a density of concrete, 𝜌𝜌c = 150
lb
ft3
, the deck contribution to the overall weight
per unit length, 𝑤𝑤c, was calculated
𝑤𝑤c = 𝜌𝜌c 𝐴𝐴c = �150
lb
ft3
� (43 ft)(1 ft) �
1 kip
1000 lb
� = 6.45
kip
ft
3
http://www.steelconstruction.info/Bridges_-_initial_design, based upon a typical highway bridge of two or
more spans carrying a single carriageway over a dual 2 or dual 3 lane highway.
PAGE 4
7. OVERALL WEIGHT
The overall weight per unit length, 𝑤𝑤, was calculated by taking the sum of the deck
contribution and the girder contribution multiplied by a factor to account for the
undetermined weight of the cross girders and bracing
𝑤𝑤 = 2𝑤𝑤st + 𝑤𝑤c = 2(1.42) + 6.45 = 9.29
kip
ft
Multiplying by the total length of the bridge gives the total weight that is to be lumped at
the deck level
𝑊𝑊 = 𝑤𝑤𝑤𝑤 = �9.29
kip
ft
� (375 ft) = 3484 kips
2. NATURAL FREQUENCY AND PERIOD
STIFFNESS
The stiffness of each column in a bent is calculated
𝑘𝑘col =
12𝐸𝐸𝐸𝐸
ℎ3
In Section 1, the column height, ℎ, and moment of inertia, 𝐼𝐼, were specified as 25 ft and
13 ft4
, respectively. The assumed elastic modulus of concrete was 𝐸𝐸 = 3000 ksi.
Substituting these values, the column stiffness is
𝑘𝑘col =
12 (3000 ksi) (13ft2)
(25 ft)3
�
144 in2
1 ft2
� = 4313
kip
ft
Each bent consists of three columns, therefore the bent stiffness, assuming the bridge
displaces rigidly as shown in Figure 2, is
𝑘𝑘bent = 3 𝑘𝑘col = 3 �4313
kip
ft
� = 12940
kip
ft
The structure has two bents, therefore the system stiffness is
𝑘𝑘 = 2𝑘𝑘bent = 2 �12940
kip
ft
� = 25800
kip
ft
PAGE 5
8. MASS
The mass lumped at the deck level is simply the weight divided by the acceleration due to
gravity. Written mathematically,
𝑚𝑚 =
𝑊𝑊
𝑔𝑔
The weight was specified at the end of Section 1 as 3484 kips. Assuming a value 𝑔𝑔 =
32.2
ft
sec2
, the system mass was computed as
𝑚𝑚 =
3484 kips
32.2
ft
sec2
= 108.2
kip sec2
ft
NATURAL FREQUENCY
The following equation shows the natural frequency, 𝜔𝜔𝑛𝑛, being a function of the two
system characteristics just presented:
𝜔𝜔𝑛𝑛 = �
𝑘𝑘
𝑚𝑚
Substituting the values for the system mass and stiffness, the natural frequency is
𝜔𝜔𝑛𝑛 = �
25800 kip/ft
108.2 kip sec2/ft
= 15.4
rad
sec
NATURAL PERIOD
The natural period, 𝜏𝜏𝑛𝑛, can easily be calculated by
𝜏𝜏𝑛𝑛 =
2𝜋𝜋
𝜔𝜔𝑛𝑛
Given the calculated natural frequency, the natural period is
𝜏𝜏𝑛𝑛 =
2𝜋𝜋
15.4 sec
= 0.41 sec
3. UNDAMPED FREE VIBRATION
The undamped free vibration response of the structure to two distinct cases is shown in
Figure 4.
PAGE 6
9. Figure 4. Free vibration response.
The initial conditions for the two distinct cases were
i. 𝑢𝑢(0) = 0.5 ft, 𝑢𝑢̇(0) = 10
ft
sec
ii. 𝑢𝑢(0) = 0.2 ft, 𝑢𝑢̇(0) = −10
ft
sec
The mathematical solution of the structure’s response in both cases is
i. 𝑢𝑢(𝑡𝑡) = 0.5 cos 15.4𝑡𝑡 +
10
15.4
sin 15.4𝑡𝑡 ft
ii. 𝑢𝑢(𝑡𝑡) = 0.2 cos 15.4𝑡𝑡 −
10
15.4
sin 15.4𝑡𝑡 ft
4. VISCOUSLY DAMPED FREE VIBRATION
In this section, the structure was assumed to have a damping ratio of 0.02, 0.05, 0.15, 0.7,
1.0 and 1.5. For each damping ratio, the natural frequency of damped vibration can be
calculated
𝜔𝜔𝐷𝐷 = 𝜔𝜔𝑛𝑛�1 − 𝜁𝜁2
PAGE 7
10. PART (I)
For the selected initial conditions in case (ii), the mathematical solution for the three
distinct cases of damped motion (underdamped, critically damped, and overdamped) is
given
1) Underdamped, 𝜁𝜁 < 1
𝑢𝑢(𝑡𝑡) = 𝑒𝑒−𝜁𝜁𝜔𝜔 𝑛𝑛 𝑡𝑡
�𝑢𝑢(0) cos 𝜔𝜔𝐷𝐷 𝑡𝑡 +
𝑢𝑢̇(0) + 𝜁𝜁𝜔𝜔𝑛𝑛 𝑢𝑢(0)
𝜔𝜔𝐷𝐷
sin 𝜔𝜔𝐷𝐷 𝑡𝑡�
2) Critically damped, 𝜁𝜁 = 1
𝑢𝑢(𝑡𝑡) = 𝑒𝑒−𝜁𝜁𝜔𝜔 𝑛𝑛 𝑡𝑡 [𝑢𝑢(0) + (𝑢𝑢̇(0) + 𝜔𝜔𝑛𝑛 𝑢𝑢(0))𝑡𝑡]
3) Overdamped, 𝜁𝜁 > 1
𝑢𝑢(𝑡𝑡) = 𝐶𝐶1 𝑒𝑒
�−𝜁𝜁+�𝜁𝜁2−1�𝜔𝜔𝑛𝑛 𝑡𝑡
+ 𝐶𝐶2 𝑒𝑒
�−𝜁𝜁−�𝜁𝜁2−1�𝜔𝜔 𝑛𝑛 𝑡𝑡
where
𝐶𝐶1 =
𝑢𝑢(0)𝜔𝜔𝑛𝑛�𝜁𝜁 + �𝜁𝜁2 − 1� + 𝑢𝑢̇(0)
2𝜔𝜔𝑛𝑛�𝜁𝜁2 − 1
𝐶𝐶2 =
−𝑢𝑢(0)𝜔𝜔𝑛𝑛�𝜁𝜁 − �𝜁𝜁2 − 1� − 𝑢𝑢̇(0)
2𝜔𝜔𝑛𝑛�𝜁𝜁2 − 1
A solution was obtained for each damping ratio with MATLAB. The results are shown in
Figure 5. The result for each underdamped case was plotted blue in color to help make a
comparison of the underdamped cases. As expected, a response of an underdamped
system shows a decrease in response amplitude with time. A second observation is the
rate at which the amplitudes decreases follows the damping ratio (i.e., the underdamped
solution with the largest damping ratio dies out the fastest). Also, it can be observed
from the response of the critically damped and overdamped system that no oscillatory
motion is present for these cases, another expected result.
PAGE 8
11. Figure 5. Damped responses to initial conditions.
PART (II)
For each underdamped case, the ratio of two consecutive amplitudes was evaluated using
𝑢𝑢𝑖𝑖
𝑢𝑢𝑖𝑖+1
= exp �
2𝜋𝜋𝜋𝜋
�1 − 𝜁𝜁2
�
and the logarithmic decrement, 𝛿𝛿, using
𝛿𝛿 =
2𝜋𝜋𝜋𝜋
�1 − 𝜁𝜁2
The following table presents the ratio and logarithmic decrement for each underdamped
case.
PAGE 9
12. Table 2. Ratio of two consecutive amplitudes
Damping ratio, 𝜻𝜻 𝒖𝒖𝒊𝒊
𝒖𝒖𝒊𝒊+ 𝟏𝟏
𝜹𝜹
0.02 1.13 0.13
0.05 1.37 0.31
0.15 2.59 0.95
0.7 472.84 6.16
5. STRUCTURE SUBJECTED TO SINUSOIDAL HORIZONTAL FORCE
The frequency of the exciting force was varied as 0.5, 1, 2 and 3 times the natural
frequency of the system. For the purposes of this project, this quantity is termed the
frequency ratio 𝑟𝑟. The amplification factor for each frequency ratio was determined using
𝑅𝑅𝑑𝑑 =
1
�(1 − 𝑟𝑟2)2 + (2𝜁𝜁𝜁𝜁)2
where
𝑟𝑟 =
𝜔𝜔
𝜔𝜔𝑛𝑛
and
𝜁𝜁 = 0.1
The following table presents the amplification factor for each frequency ratio.
Table 3. Amplification factors.
𝒓𝒓 𝑹𝑹𝒅𝒅
0.5 1.32
1.0 5.0
2.0 0.33
3.0 0.12
PAGE 10
13. 6. LIMITS FOR FREQUENCY OF VIBRATING MACHINE
The limit for the frequency of a vibrating machine installed on the deck of the bridge was
determined so that the amplitude of the transmitted force to the deck is 20% of the
exciting amplitude. Stated differently, for a given damping ratio, the frequency at which
the transmissibility, 𝑇𝑇𝑇𝑇, is equal to 0.2 was determined. Transmissibility vs. frequency
ratio is shown in Figure 6. The red squares show the intersections of each curve with the
line 𝑇𝑇𝑇𝑇 = 0.2. The frequency at each of these locations is presented in Table 4.
Figure 6. Transmissibility Plot.
Table 4. Frequencies for TR = 0.2.
𝜻𝜻 𝝎𝝎 [𝐇𝐇𝐇𝐇]
0.02 6.03
0.05 6.09
0.15 6.69
0.70 17.30
PAGE 11
14. In summary, for a lightly damped, 𝜁𝜁 ≤ 0.15, vibrating machine, the excited frequency
should be
𝜔𝜔 > 6.69 Hz
For a significantly damped, 𝜁𝜁 ≤ 0.7, machine, the exciting frequency should be
𝜔𝜔 > 17.30 Hz
7. PERIODIC RECTANGULAR WAVE
A periodic rectangular wave (plotted black in color in the following four figures) with unit
amplitude and period 𝜏𝜏 was solved in class. The period was selected as 2 sec for the
project. From observation of the results presented in Figure 7 - Figure 10, seven Fourier
terms are needed to sufficiently reproduce its shape. No noticeable difference is seen after
increasing the number of terms from seven to eight, therefore seven was the number
chosen.
Figure 7. Fourier series with n = 1 and n = 2 terms.
PAGE 12
17. Figure 10. Fourier series with n = 7 and n = 8 terms.
8. STRUCTURAL RESPONSE TO ARBITRARY EXCITATION
The arbitrary excitation selected for the project was
𝐹𝐹(𝑡𝑡) = 𝐹𝐹0(1 − 𝑡𝑡2) + 𝐹𝐹1 sin 𝜔𝜔1 𝑡𝑡 + 𝐹𝐹2 sin 𝜔𝜔2 𝑡𝑡
where
𝐹𝐹0 = 15 kip, 𝐹𝐹1 = 4 kip, 𝐹𝐹2 = 2 kip, 𝜔𝜔1 = 18
rad
sec
and 𝜔𝜔2 = 59
rad
sec
This function is shown in Figure 11.
PAGE 15
18. Figure 11. Arbitrary excitation.
The response of the bridge to this excitation was determined by numerically integrating
Duhamel’s integral with MATLAB. A damping ratio of 0.1 was assumed for the bridge.
The excitation was applied for a duration of 0.8 sec and the response is shown in Figure
12; the portion of the response during the excitation is plotted in blue and the portion
after in orange.
PAGE 16
19. Figure 12. Response to arbitrary excitation.
CONCLUSION
The CIVE 801 course project was an alternative way to demonstrate knowledge of
dynamics of SDF systems. A significant effort was made to be thorough and clear in the
presentation of the work completed to fulfill the requirements of the project. From a
student’s perspective, the author considers the project more beneficial, and much more
enjoyable, than studying for and taking a final exam. All parts of the project, except for
the frame design, were completed in MATLAB. The appendix contains the MATLAB code
for the project.
PAGE 17
20. APPENDIX
This appendix was published with MATLAB® R2014b
1. SYSTEM CHARACTERISTICS
Stiffness, [kip/ft]
k = 25800;
Mass, [kip-sec2/ft]
m = 108.2;
2. DETERMINE NATURAL FREQUENCY AND PERIOD OF STRUCTURE
Natural frequency, [rad/sec]
w_n = sqrt(k/m)
w_n =
15.4417
Natural period, [sec]
tau_n = 2*pi/w_n
tau_n =
0.4069
3. SUBJECT STRUCTURE TO INITIAL CONDITIONS
Mathematical solution for free vibration without damping
syms u0 v0 t
u_nodamp = u0*cos(w_n*t)+(v0/w_n)*sin(w_n*t);
t = linspace(0,3*tau_n);
PAGE 18
21. Case (i):
u0 = 0.5; % [ft]
v0 = 10; % [ft/s]
u_i = double(subs(u_nodamp));
Case (ii):
u0 = 0.2; % [ft]
v0 = -10; % [ft/s]
u_ii = double(subs(u_nodamp));
Plot the response from case (i) and case (ii)
figure(1)
plot(t,u_i,t,u_ii)
xlabel('t (s)')
ylabel('u (ft)')
legend('Case (i)','Case (ii)')
4. ASSUME DAMPING RATIOS OF 2%, 5%, 15%, 70%, 100% AND 150%
4(i)
Mathematical solution for underdamped, , cases
syms zeta t
w_D = w_n*sqrt(1-zeta^2);
u_underDamp = exp(-zeta*w_n*t)*(u0*cos(w_D*t)+...
(v0+zeta*w_n*u0)*sin(w_D*t)/w_D);
Mathematical solution for critically damped, , case
u_critDamp = exp(-w_n*t)*(u0+(v0+w_n*u0)*t);
Mathematical solution for overdamped, , case
C1 = (u0*w_n*(zeta+sqrt(zeta^2-1))+v0)/(2*w_n*sqrt(zeta^2-1));
C2 = (-u0*w_n*(zeta-sqrt(zeta^2-1))-v0)/(2*w_n*sqrt(zeta^2-1));
u_overDamp = C1*exp((-zeta+sqrt(zeta^2-1))*w_n*t)+...
C2*exp((-zeta-sqrt(zeta^2-1))*w_n*t);
PAGE 19
23. Evaluate
𝑢𝑢𝑖𝑖
𝑢𝑢𝑖𝑖+1
at underdamped cases
for i = 1:length(zeta_vec)
ratios(i) = r(zeta_vec(i));
decrements(i) = ln_r(zeta_vec(i));
end
Print table
fprintf('%15s%15s%25sn','damping ratio','u_i/u_(i+1)','logarithmic decrement')
fprintf('%15s%15s%25sn','-------------','-----------','---------------------')
for i = 1:length(ratios)
fprintf('%15.2f%15.2f%25.2fn',zeta_vec(i),ratios(i),decrements(i))
end
fprintf('n')
damping ratio u_i/u_(i+1) logarithmic decrement
------------- ----------- ---------------------
0.02 1.13 0.13
0.05 1.37 0.31
0.15 2.59 0.95
0.70 472.84 6.16
5. SUBJECT STRUCTURE TO A SINUSOIDAL HORIZONTAL FORCE
Assume damping ratio
zeta = 0.1;
Expression for amplification factor
R_d = @(r) 1/sqrt((1-r^2)^2+(2*zeta*r)^2);
Vector containing values of
𝜔𝜔
𝜔𝜔 𝑛𝑛
at which 𝑅𝑅𝑑𝑑 will be evaluated
r_values = [0.5 1 2 3];
Evaluate
PAGE 21
24. for i = 1:length(r_values)
Rd_values(i) = R_d(r_values(i));
end
Table
fprintf('%24s%24sn','Frequency ratio (w/w_n)','Amplification Factor')
fprintf('%24s%24sn','-----------------------','--------------------')
for i = 1:length(r_values)
fprintf('%24.1f%24.2fn',r_values(i),Rd_values(i))
end
fprintf('n')
Frequency ratio (w/w_n) Amplification Factor
----------------------- --------------------
0.5 1.32
1.0 5.00
2.0 0.33
3.0 0.12
6. LIMITS FOR THE FREQENCY OF VIBRATING MACHINE FOR TR = 20%
CONDITION
Expression for transmissibility, TR
syms r positive
syms zeta
TR = sqrt((1+(2*zeta*r)^2)/((1-r^2)^2+(2*zeta*r)^2));
Output vector for ratios and forcing frequencies that satisfy TR condition
r = zeros(1,length(zeta_vec));
w = zeros(1,length(zeta_vec));
Results
for i = 1:length(zeta_vec)
[S.r,~]=solve(TR==0.2,zeta==zeta_vec(i),'Real',true,...
'IgnoreAnalyticConstraints', true);
r(i) = double(S.r);
w(i) = r(i)*w_n/(2*pi);
end
TR = @(r,zeta) sqrt((1+(2*zeta*r)^2)/((1-r^2)^2+(2*zeta*r)^2));
r_plot = linspace(0,8,200);
PAGE 22
25. TR_plot = zeros(length(zeta_vec),length(r_plot));
for i = 1:length(zeta_vec)
for j = 1:length(r_plot)
TR_plot(i,j) = TR(r_plot(j),zeta_vec(i));
end
end
Table
fprintf('%15s%31sn','damping ratio','frequency for TR = 0.2 [Hz]')
fprintf('%15s%31sn','-------------','---------------------------')
for i = 1:length(w)
fprintf('%15.2f%31.2fn',zeta_vec(i),w(i))
end
fprintf('n')
damping ratio frequency for TR = 0.2 [Hz]
------------- ---------------------------
0.02 6.03
0.05 6.09
0.15 6.69
0.70 17.30
Plot
figure(3)
plot(r_plot,TR_plot(1,:),...
r_plot,TR_plot(2,:),...
r_plot,TR_plot(3,:),...
r_plot,TR_plot(4,:),...
[r_plot(1) r_plot(length(r_plot))],[0.2 0.2],'k:',...
[r_plot(1) r_plot(length(r_plot))],[1 1],'k-',...
[1 1],[0 15],'k-',...
r(1),TR(r(1),zeta_vec(1)),'rs',...
r(4),TR(r(4),zeta_vec(4)),'rs')
ylabel('TR')
xlabel('omega/omega_n')
axis([0 8 0 15])
set(gca,'YTick',[0.2,1:15])
legend('zeta = 0.02',...
'zeta = 0.05',...
'zeta = 0.15',...
'zeta = 0.7')
7. FOURIER REPRESENTATION OF PERIODIC RECTANGULAR WAVE
Rectangular wave parameters
PAGE 23
26. Fo = 2; % [kip]
tau = 2; % [s]
Data for rectangular wave plot
t_top = -2*tau:tau/2:3*tau/2;
t_bot = -3*tau/2:tau/2:2*tau;
x_top = (Fo/2)*ones(1,8);
x_bot = -1*x_top;
Fourier Series
f = @(t,n) (2*Fo/(pi*n))*sin(2*pi*n*t/tau);
Number of terms for approximation
num = 8;
Approximations
f_approx = 0;
t_values = linspace(-2*tau,2*tau,200);
f_values = zeros(num,length(t_values));
for i = 1:num
syms t
n = 2*i-1;
f_approx = f_approx + f(t,n);
for j = 1:length(t_values)
t = t_values(j);
f_values(i,j) = double(subs(f_approx));
end
end
Plot
for i = 1:4
figure
% Rectangular wave
plot(t_top(1:2),x_top(1:2),'k-',t_bot(1:2),x_bot(1:2),'k-')
hold on
plot(t_top(3:4),x_top(3:4),'k-',t_bot(3:4),x_bot(3:4),'k-')
plot(t_top(5:6),x_top(5:6),'k-',t_bot(5:6),x_bot(5:6),'k-')
plot(t_top(7:8),x_top(7:8),'k-',t_bot(7:8),x_bot(7:8),'k-')
for j = 2:8
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27. plot([t_top(j) t_top(j)],[1 -1],'k--')
end
axis([-2*tau 2*tau -2 2])
xlabel('Time (s)')
ylabel('Force (kip)')
set(gca,'YTick',-2:2)
% Fourier approximations
plot(t_values,f_values(2*i-1,:))
plot(t_values,f_values(2*i,:))
legend(['n = ' num2str(2*i-1)],...
['n = ' num2str(2*i)])
end
8. RESPONSE OF STRUCTURE TO ARBITRARY EXCITATION
Excitation function
F0 = 15; % [kip]
F1 = 4; % [kip]
F2 = 2; % [kip]
w1 = 18; % [rad/s]
w2 = 59; % [rad/s]
F = @(x) F0.*(1-x.^2)+F1.*sin(w1.*x)+F2.*sin(w2.*x);
Plot function
t = linspace(0,0.8);
F_values = zeros(1,length(t));
for i = 1:length(t)
F_values(i) = F(t(i));
end
figure
plot(t,F_values)
xlabel('Time (s)')
ylabel('Force (kip)')
Unit impulse-response function
zeta = 0.1;
w_D = w_n*sqrt(1-zeta^2);
h = @(x,t) -(1/w_D).*exp(-zeta*w_n.*(t-x)).*sin(w_D.*(t-x));
Function for Duhamel's integral
fun = @(x,t) F(x).*h(x,t);
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28. Response during excitation
t1 = linspace(0,0.8);
u1 = zeros(1,length(t1));
for i=1:length(t1)
u1(i) = integral(@(tau)fun(tau,t1(i)),0,t1(i));
end
Response after excitation
idx = length(u1);
u0 = u1(idx);
v0 = (u1(idx)-u1(idx-2))/(t1(idx)-t1(idx-2));
t2 = linspace(0,4);
u_underDamp = @(t) exp(-zeta*w_n.*t).*(u0*cos(w_D.*t)+...
(v0+zeta*w_n*u0).*sin(w_D.*t)/w_D);
u2 = u_underDamp(t2);
figure
plot(t1,u1,t2+0.8,u2)
xlabel('Time (s)')
ylabel('Displacment (ft)')
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