cive 801 - course project report (teal)

CIVE 801 DYNAMICS OF STRUCTURES I COURSE PROJECT

Cyle Teal CIVE 801, Winter 14-15 March 16, 2015

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

7. Fourier representation of periodic rectangular wave ................................................. 23

8. Response of structure to arbitrary excitation ............................................................. 25

PAGE 1

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

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

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 lbin3

, the longitudinal girder contribution to the overall weight per unit length was calculated

𝑤𝑤g = 4 𝜌𝜌st 𝐴𝐴st = 4 �0.28lb

in3� (105.3 in2) �

1 kip1000 lb

� �12 in1 ft

� = 1.42 kipft

DECK WEIGHT

Assuming a density of concrete, 𝜌𝜌c = 150 lbft3

, the deck contribution to the overall weight

per unit length, 𝑤𝑤c, was calculated

𝑤𝑤c = 𝜌𝜌c 𝐴𝐴c = �150lbft3� (43 ft)(1 ft) �

1 kip1000 lb

� = 6.45 kipft

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

http://www.steelconstruction.info/Bridges_-_initial_design

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 kipft

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 kipft� (375 ft) = 3484 kips

2. NATURAL FREQUENCY AND PERIOD

STIFFNESS

The stiffness of each column in a bent is calculated

𝑘𝑘col =12𝐸𝐸𝐼𝐼ℎ3

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

kipft

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 kipft� = 12940

kipft

The structure has two bents, therefore the system stiffness is

𝑘𝑘 = 2𝑘𝑘bent = 2 �12940 kipft� = 25800

kipft

PAGE 5

MASS

The mass lumped at the deck level is simply the weight divided by the acceleration due to gravity. Written mathematically,

𝑚𝑚 =𝑊𝑊𝑔𝑔

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 ftsec2

= 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 = �𝑘𝑘𝑚𝑚

Substituting the values for the system mass and stiffness, the natural frequency is

𝜔𝜔𝑛𝑛 = �25800 kip/ft

108.2 kip sec2/ft= 15.4

radsec

NATURAL PERIOD

The natural period, 𝜏𝜏𝑛𝑛, can easily be calculated by

𝜏𝜏𝑛𝑛 =2𝜋𝜋𝜔𝜔𝑛𝑛

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

Figure 4. Free vibration response.

The initial conditions for the two distinct cases were

i. 𝑢𝑢(0) = 0.5 ft, �̇�𝑢(0) = 10 ftsec

ii. 𝑢𝑢(0) = 0.2 ft, �̇�𝑢(0) = −10 ftsec

The mathematical solution of the structure’s response in both cases is

i. 𝑢𝑢(𝑡𝑡) = 0.5 cos 15.4𝑡𝑡 + 1015.4

sin 15.4𝑡𝑡 ft

ii. 𝑢𝑢(𝑡𝑡) = 0.2 cos 15.4𝑡𝑡 − 1015.4

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

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−𝜁𝜁𝜔𝜔𝑛𝑛𝑡𝑡 �𝑢𝑢(0) cos𝜔𝜔𝐷𝐷𝑡𝑡 +�̇�𝑢(0) + 𝜁𝜁𝜔𝜔𝑛𝑛𝑢𝑢(0)

𝜔𝜔𝐷𝐷sin𝜔𝜔𝐷𝐷𝑡𝑡�

2) Critically damped, 𝜁𝜁 = 1 𝑢𝑢(𝑡𝑡) = 𝑒𝑒−𝜁𝜁𝜔𝜔𝑛𝑛𝑡𝑡 [𝑢𝑢(0) + (�̇�𝑢(0) + 𝜔𝜔𝑛𝑛𝑢𝑢(0))𝑡𝑡]

3) Overdamped, 𝜁𝜁 > 1

𝑢𝑢(𝑡𝑡) = 𝐶𝐶1𝑒𝑒�−𝜁𝜁+�𝜁𝜁2−1�𝜔𝜔𝑛𝑛𝑡𝑡 + 𝐶𝐶2𝑒𝑒

�−𝜁𝜁−�𝜁𝜁2−1�𝜔𝜔𝑛𝑛𝑡𝑡 where

𝐶𝐶1 =𝑢𝑢(0)𝜔𝜔𝑛𝑛�𝜁𝜁 + �𝜁𝜁2 − 1� + �̇�𝑢(0)

2𝜔𝜔𝑛𝑛�𝜁𝜁2 − 1

𝐶𝐶2 =−𝑢𝑢(0)𝜔𝜔𝑛𝑛�𝜁𝜁 − �𝜁𝜁2 − 1� − �̇�𝑢(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

Figure 5. Damped responses to initial conditions.

PART (II)

For each underdamped case, the ratio of two consecutive amplitudes was evaluated using

𝑢𝑢𝑖𝑖𝑢𝑢𝑖𝑖+1

= exp�2𝜋𝜋𝜁𝜁

�1 − 𝜁𝜁2�

and the logarithmic decrement, 𝛿𝛿, using

𝛿𝛿 =2𝜋𝜋𝜁𝜁

�1 − 𝜁𝜁2

The following table presents the ratio and logarithmic decrement for each underdamped case.

PAGE 9

Table 2. Ratio of two consecutive amplitudes

Damping ratio, 𝜻𝜻 𝒖𝒖𝒊𝒊𝒖𝒖𝒊𝒊+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 + (2𝜁𝜁𝑟𝑟)2

where

𝑟𝑟 =𝜔𝜔𝜔𝜔𝑛𝑛

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

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, 𝑇𝑇𝑅𝑅, 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 𝑇𝑇𝑅𝑅 = 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

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


PAGE 13


PAGE 14


8. STRUCTURAL RESPONSE TO ARBITRARY EXCITATION The arbitrary excitation selected for the project was

𝐹𝐹(𝑡𝑡) = 𝐹𝐹0(1 − 𝑡𝑡2) + 𝐹𝐹1 sin𝜔𝜔1𝑡𝑡 + 𝐹𝐹2 sin𝜔𝜔2𝑡𝑡

where

𝐹𝐹0 = 15 kip, 𝐹𝐹1 = 4 kip, 𝐹𝐹2 = 2 kip, 𝜔𝜔1 = 18 radsec

and 𝜔𝜔2 = 59 radsec

This function is shown in Figure 11.

PAGE 15

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

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

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

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

Plot results

zeta_vec = [0 .02 .05 .15 .7 1 1.5];

t_vec = linspace(0,6*tau_n,200)';

u_results = zeros(length(t),length(zeta));

for j = 1:length(zeta_vec)

zeta = zeta_vec(j);

for k = 1:length(t_vec)

t = t_vec(k);

if zeta < 1

u_results(k,j) = double(subs(u_underDamp));

elseif zeta == 1

u_results(k,j) = double(subs(u_critDamp));

else

u_results(k,j) = double(subs(u_overDamp));

end

end

end

figure(2)

plot(t_vec,u_results(:,1),'k',...

t_vec,u_results(:,2),'b--',...

t_vec,u_results(:,3),'b:',...

t_vec,u_results(:,4),'b-.',...

t_vec,u_results(:,5),'b-',...

t_vec,u_results(:,6),'r',...

t_vec,u_results(:,7),'g')

legend('\zeta = 0',...

'\zeta = 0.02',...

'\zeta = 0.05',...

'\zeta = 0.15',...

'\zeta = 0.7',...

'\zeta = 1',...

'\zeta = 1.5',...

'Location','NorthEastOutside')

axis([0 6*tau_n -1 1])

set(gca,'XTick',0.5:0.5:2,'YTick',-1:0.5:1)

text(.6,.8,'ICs: u(0) = 0.2 ft, v(0) = -10 ft/s')

xlabel('Time (s)')

ylabel('Displacement (ft)')

4(ii)

Expression for ratio two consecutive amplitudes

r = @(zeta) exp(2*pi*zeta/sqrt(1-zeta^2));

ln_r = @(zeta) 2*pi*zeta/sqrt(1-zeta^2);

zeta_vec = [.02 .05 .15 .7];

PAGE 20

Evaluate 𝑢𝑢𝑖𝑖𝑢𝑢𝑖𝑖+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%25s\n','damping ratio','u_i/u_(i+1)','logarithmic decrement')

fprintf('%15s%15s%25s\n','-------------','-----------','---------------------')

for i = 1:length(ratios)

fprintf('%15.2f%15.2f%25.2f\n',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 𝜔𝜔𝜔𝜔𝑛𝑛

at which 𝑅𝑅𝑑𝑑 will be evaluated

r_values = [0.5 1 2 3];

Evaluate

PAGE 21

for i = 1:length(r_values)

Rd_values(i) = R_d(r_values(i));

end

Table

fprintf('%24s%24s\n','Frequency ratio (w/w_n)','Amplification Factor')

fprintf('%24s%24s\n','-----------------------','--------------------')

for i = 1:length(r_values)

fprintf('%24.1f%24.2f\n',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


[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

TR_plot = zeros(length(zeta_vec),length(r_plot));


for j = 1:length(r_plot)

TR_plot(i,j) = TR(r_plot(j),zeta_vec(i));

end

end

Table

fprintf('%15s%31s\n','damping ratio','frequency for TR = 0.2 [Hz]')

fprintf('%15s%31s\n','-------------','---------------------------')

for i = 1:length(w)

fprintf('%15.2f%31.2f\n',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(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

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




for j = 2:8

PAGE 24

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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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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cive 801 - course project report (teal)

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