Matlab programs

2/7/2016 HW_2_Q1
file:///E:/important%20%20books/structural%20dynamcis/results/html/HW_2_Q1.html 1/2
%equation of motion = P0*cos(omega*t)= m*d2u + k*u
% the response function = (U0/1‐r^2)*cos(omega*t) + A1*cos(wn*t)+ A2
% *sin(wn*t)
% for M = 2,k=32,F = P*cos(omega*t)[P =16 ;omega = 4.1],wn
% =sqrt(k/M0); solving the equation and plotting the graph we get
t = linspace(0,235.61,10000);%given time intervals = 150
u = ‐9.876 *cos(4.1.*t) + 9.876 * cos(4.*t) ;%equation got from the response function
figure
plot (t,u,'b‐')
xlabel ('time')
ylabel ('displacement')
title ('beating solution cosine input')
legend ( 'numerical solution')
grid on
Published with MATLAB® R2015a

2/7/2016 HW_2_Q1

2/7/2016 HW_2_Q2
% for a resonance problem we assume that up(particular solution is of the
% following
%up = U0*wn*t/2 *sin(wn*t) + A1* cos(wn*t) + A2 * sin (wn * t)
%for m =2, k = 32 , p = 16 , wn = omega = 4 ;
% the response function is as follows
% u = t*sin(4*t);
clc
clear all
t =linspace(0,235.61,10000); % Tn = 2*pi/wn , t = n*Tn
u = t .* sin(4.*t);
figure
plot ( t,u,'r‐')
xlabel ('time')
title ('resonant solution')
legend ('numerical solution')
grid on

2/7/2016 HW_2_Q2

2/7/2016 HW2_Q3
file:///E:/important%20%20books/structural%20dynamcis/results/html/HW2_Q3.html 1/4
load HW2_Prob_3_Data.mat% saved data
time=SOL(:,1)
U=SOL(:,2)
figure
plot(time,U)
grid on
xlabel ('time')
title ('free vibration f damped system(using saved data)')
%by the method of logarithmic decreament
% at time =  0.9322 up=0.4473
%at time = 1.9492 uq = 0.3069
% formula z/sqrt(1‐z^2).* 2*pi = log(up/uq)
%since z <<1 , formula = 2*pi*z = log(up/uq)
up = 0.4473;
uq = 0.3069;
z = (log(up/uq))/(2*pi)
time =
         0
    0.0847
    0.1695
    0.2542
    0.3390
    0.4237
    0.5085
    0.5932
    0.6780
    0.7627
    0.8475
    0.9322
    1.0169
    1.1017
    1.1864
    1.2712
    1.3559
    1.4407
    1.5254
    1.6102
    1.6949
    1.7797
    1.8644
    1.9492
    2.0339
    2.1186
    2.2034
    2.2881
    2.3729
    2.4576
    2.5424
    2.6271
    2.7119

2/7/2016 HW2_Q3
    2.7966
    2.8814
    2.9661
    3.0508
    3.1356
    3.2203
    3.3051
    3.3898
    3.4746
    3.5593
    3.6441
    3.7288
    3.8136
    3.8983
    3.9831
    4.0678
    4.1525
    4.2373
    4.3220
    4.4068
    4.4915
    4.5763
    4.6610
    4.7458
    4.8305
    4.9153
    5.0000
U =
         0
    0.1628
    0.3007
    0.3994
    0.4494
    0.4473
    0.3955
    0.3017
    0.1782
    0.0400
   ‐0.0970
   ‐0.2175
   ‐0.3087
   ‐0.3617
   ‐0.3723
   ‐0.3409
   ‐0.2728
   ‐0.1772
   ‐0.0658
    0.0484
    0.1524
    0.2351
    0.2882
    0.3069
    0.2907
    0.2428

2/7/2016 HW2_Q3
    0.1698
    0.0809
   ‐0.0133
   ‐0.1022
   ‐0.1759
   ‐0.2270
   ‐0.2507
   ‐0.2454
   ‐0.2131
   ‐0.1583
   ‐0.0881
   ‐0.0111
    0.0640
    0.1288
    0.1766
    0.2028
    0.2052
    0.1847
    0.1445
    0.0898
    0.0274
   ‐0.0355
   ‐0.0918
   ‐0.1356
   ‐0.1624
   ‐0.1700
   ‐0.1583
   ‐0.1295
   ‐0.0875
   ‐0.0374
    0.0147
    0.0631
    0.1024
    0.1286
z =
    0.0600

2/7/2016 HW2_Q3

2/7/2016 HW2_Q4
m=2;
k=32;
c =0.5;
p=16;
i = sqrt(‐1);
%omega = forcing frequency
wn = 4;
omega =linspace(0,4*wn,2000);
U= p./(k‐omega.^2.*m+i.*omega*c);
v = i.*omega.*U;
a = ‐omega.*2.*U;
figure
%nyquest polar
plot(real(U),imag(U),'r‐')
hold on
plot(real(v),imag(v),'b‐')
hold on
plot(real(a),imag(a),'g‐')
ylabel ('imaginary response')
xlabel ('real response')
grid on
title ('nyquest plot')
axis equal
figure
%co quad plot
plot ( omega , real(U),'r.')
hold on
plot ( omega , imag(U),'b.')
xlabel ('forcing frequency')
ylabel ('response')
title ('forcing frequency vs response')
legend ('real response','imaginary response')
figure
plot ( omega ,real (v),'r.')
hold on
plot (omega, imag(v),'b.')
ylabel('velocitty')
title ('forcing frequrncy vs velocity')
figure
plot(omega,real(a),'r.')
hold on
plot(omega,imag(a),'b.')
ylabel ('acceleration')
title ('acceleration vs forcing frequency')
grid on

2/7/2016 HW2_Q4

clear
clc
syms y t
u0=0;
v0=0;
P0=2;
Wn=5;
m=2;
M=pi/2;
time = linspace ( 0,1,200);
U1 = u0* cos(Wn.*time)+v0/Wn* sin(Wn.*time);
plot(time,U1,'k')
hold on
tt = t -1;
yy = y -1 ;
hy=(1/m*Wn)*sin(Wn*((tt)-(yy)));
fy=2*sin(M*(yy));
u2=int(hy*fy,y,0,tt);
v2=diff(u2,t);
T=linspace(1,3,200);
U2 = double (subs(u2,t,T));
V2 = double(subs(v2,t,T));
plot(T,U2,'r-');
hold on
U=double(subs(u2,t,3));
V=double(subs(v2,t,3));
t=linspace(3,8,200);
tt1 = t - 3 ;
Dis=U.*cos(Wn.*(tt1))+(V/Wn).*sin(Wn.*(tt1));
plot(t,Dis,'b-')
xlabel ('Time')
ylabel ('Displacement')
title ( 'Delayed sine force response')
legend ( 'Initial response','Forced response','Residual free vibration')

clear all
clc
syms y t
u0=0;
v0=0;
Wn=5;
m=2;
plot(time,U1,'k')
hold on
tt = t -5;
yy = y -5 ;
hy=(1/m*Wn)*sin(Wn*((tt)-(yy)));
fy=1.2;
v2=diff(u2,t);
U2 = double (subs(u2,t,T))
plot(T,U2,'r-');
hold on
U=double(subs(u2,t,7))
tt1 = t - 7 ;
plot(t,Dis,'b-')
hold on
xlabel ('Time')
title ( 'Delayed rectangular pulse response')
legend ( 'Initial response','Forced response','Residual free vibration','location','Sout
hOutside' )
U2 =
Columns 1 through 7
0 -0.0032 -0.0050 -0.0052 -0.0039 -0.0011 0.0032
Columns 8 through 14
0.0091 0.0163 0.0251 0.0353 0.0469 0.0599 0.0742

0.0898 0.1068 0.1249 0.1442 0.1647 0.1863 0.2089
0.2324 0.2569 0.2823 0.3084 0.3352 0.3627 0.3908
0.4194 0.4485 0.4779 0.5076 0.5375 0.5676 0.5977
0.6279 0.6579 0.6878 0.7175 0.7468 0.7758 0.8043
0.8322 0.8596 0.8863 0.9123 0.9375 0.9618 0.9851
1.0075 1.0289 1.0491 1.0682 1.0862 1.1028 1.1182
1.1323 1.1450 1.1563 1.1662 1.1747 1.1817 1.1872
1.1912 1.1937 1.1947 1.1942 1.1922 1.1886 1.1836
1.1771 1.1691 1.1596 1.1487 1.1365 1.1228 1.1079
1.0916 1.0741 1.0554 1.0355 1.0145 0.9924 0.9693
0.9453 0.9204 0.8947 0.8682 0.8410 0.8132 0.7849
0.7561 0.7268 0.6973 0.6674 0.6374 0.6073 0.5772
0.5471 0.5171 0.4873 0.4578 0.4286 0.3999 0.3716

0.3439 0.3169 0.2905 0.2649 0.2401 0.2163 0.1934
0.1715 0.1506 0.1309 0.1124 0.0951 0.0790 0.0643
0.0509 0.0388 0.0282 0.0190 0.0112 0.0049 0.0001
-0.0031 -0.0049 -0.0052 -0.0040 -0.0012 0.0031 0.0088
0.0161 0.0248 0.0349 0.0464 0.0594 0.0737 0.0893
0.1061 0.1242 0.1435 0.1640 0.1855 0.2081 0.2316
0.2560 0.2813 0.3074 0.3343 0.3617 0.3898 0.4184
0.4474 0.4768 0.5065 0.5364 0.5665 0.5967 0.6268
0.6568 0.6867 0.7164 0.7458 0.7747 0.8033 0.8313
0.8587 0.8854 0.9114 0.9366 0.9609 0.9843 1.0068
1.0282 1.0484 1.0676 1.0855 1.1023 1.1177 1.1318
1.1446 1.1559 1.1659 1.1744 1.1814 1.1870 1.1911
1.1936 1.1947 1.1942 1.1923 1.1888 1.1838 1.1773

1.1694 1.1600 1.1492 1.1369
U =
1.1369

clear all
clc
syms y t
u0=0;
v0=0;
Wn=5;
m=2;
plot(time,U1,'k')
hold on
tt = t -5;
yy = y -5 ;
hy=0.1.*sin(5*(tt-yy));
fy=1.2;
v2=diff(u2,t);
plot(T,U2,'r-');
hold on
U=double(subs(u2,t,7))
V=double(subs(v2,t,7))
tt1 = t - 7 ;
plot(t,Dis,'b-')
hold on
clc
clear
M = 2;
K = 50;
u0 =0;
v0 = 0;
Wn =5 ;
T = linspace( 0,12,200);
syms u(t) t
du = diff(u,t);
u = dsolve ( M*diff(du,t)+ K*u==0,u(0)==0,du(0)==0);
v = diff(u,t);
time = linspace(0,5,200);
UQ = double (subs(u,t,time))
VQ = double(subs(u,t,time));

U1 = double (subs(u,t,5));
V1 = double (subs(v,t,5));
plot(time,UQ,'k*')
hold on
syms u1(t1) t1
du1 = diff(u1,t1);
u1 = dsolve ( M*diff(du1,t1)+ K*u1==1.2,u1(5)==U1,du1(5)==V1);
v1 = diff(u1,t1);
U2 = double (subs(u1,t1,time));
V2 = double(subs(u1,t1,time));
U22 = double (subs(u1,t1,7));
V22 = double (subs(v1,t1,7));
plot(time,U2,'g*')
hold on
syms u2(t2) t2
du2=diff(u2,t2);
u2=dsolve(M*diff(du2,t2)+K*u2==0,u2(7)==U22,du2(7)==V22);
v2=diff(u2,t2);
U3 = double(subs(u2,t2,time));
U4 = double (subs(v2,t2,time));
plot(time,U3,'r*')
grid on
hold on
xlabel ('Time')
legend ( 'Initial response','Forced response','Residual free vibration','location','Sout
hOutside' )
U =
0.0455
V =
-0.0514
UQ =
0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0

clear
clc
syms y t
u0=0.1;
v0=-0.5;
P0=2;
Wn=5;
m=2;
M=pi/4;
U1 = u0* cos(Wn.*t)+v0/Wn* sin(Wn.*t);
V1 = diff(U1,t);
UT = double(subs(U1,t,time));
VT = double(subs(V1,t,time));
UT1 = UT(end)
VT1 = VT(end)
plot(time,UT,'k')
hold on
tt = t -1;
yy = y -1 ;
hy=(1/m*Wn)*sin(Wn*((tt)-(yy)))
fy=2*sin(M*(yy));
u2=int(hy*fy,y,0,tt)++ UT1* cos(Wn.*tt)+VT1/Wn* sin(Wn.*tt);
v2=diff(u2,t);
plot(T,U2,'r-');
hold on
U=double(subs(u2,t,3));
tt5 = t - 3 ;
Z=U.*cos(Wn.*(tt5))+(V/Wn).*sin(Wn.*(tt5))
plot(t,Z,'b-')
hold on
%--------------------------------------------------------------------------%
syms t Z
tt5 = t - 3 ;
Z=U.*cos(Wn.*(tt5))+(V/Wn).*sin(Wn.*(tt5));
vel = diff(Z,t);
time=linspace(3,5,200);
U3 = double (subs(Z,t,time));
V3 = double (subs (vel,t,time));

U4 = U3(end)
V4 = V3(end)
syms y t
tt4 = t-5;
yy2 = y -5 ;
hy=(1/m*Wn)*sin(Wn*((tt4)-(yy2)));
fy=1.2;
u3=int(hy*fy,y,0,tt4) +U4.* cos(Wn.*tt4)+V4./Wn* sin(Wn.*tt4);
v3=diff(u3,t);
U5 = double (subs(u3,t,T)) ;
plot(T,U5,'r-');
hold on
UI=double(subs(u3,t,7));
VI=double(subs(v3,t,7));
tt5 = t - 7 ;
Dis=UI.*cos(Wn.*(tt5))+(VI/Wn).*sin(Wn.*(tt5));
plot(t,Dis,'b-')
xlabel ('Time')
legend ( 'Initial response-1','Forced response-1','Residual free vibration-1','Forced re
sponse-2','Residual free vibration-2','location','southOutside')
UT1 =
0.1243
VT1 =
0.3376
hy =
(5*sin(5*t - 5*y))/2
Z =
Columns 1 through 7
-0.6694 -0.6848 -0.6986 -0.7105 -0.7207 -0.7290 -0.7356

-0.7402 -0.7430 -0.7439 -0.7429 -0.7401 -0.7354 -0.7288
-0.7204 -0.7102 -0.6981 -0.6844 -0.6688 -0.6516 -0.6328
-0.6124 -0.5904 -0.5669 -0.5420 -0.5157 -0.4881 -0.4593
-0.4293 -0.3983 -0.3662 -0.3332 -0.2994 -0.2648 -0.2296
-0.1937 -0.1574 -0.1207 -0.0837 -0.0464 -0.0091 0.0283
0.0656 0.1027 0.1396 0.1761 0.2122 0.2478 0.2827
0.3169 0.3503 0.3828 0.4144 0.4449 0.4743 0.5025
0.5294 0.5550 0.5792 0.6019 0.6231 0.6427 0.6607
0.6770 0.6917 0.7046 0.7157 0.7250 0.7324 0.7380
0.7418 0.7437 0.7437 0.7418 0.7380 0.7324 0.7250
0.7157 0.7046 0.6917 0.6771 0.6608 0.6428 0.6231
0.6019 0.5792 0.5550 0.5295 0.5026 0.4744 0.4450
0.4145 0.3829 0.3504 0.3170 0.2828 0.2479 0.2123

0.1763 0.1397 0.1029 0.0657 0.0284 -0.0090 -0.0463
-0.0835 -0.1206 -0.1573 -0.1936 -0.2294 -0.2647 -0.2993
-0.3331 -0.3661 -0.3982 -0.4292 -0.4592 -0.4880 -0.5156
-0.5419 -0.5668 -0.5903 -0.6123 -0.6327 -0.6516 -0.6688
-0.6843 -0.6981 -0.7101 -0.7204 -0.7288 -0.7354 -0.7401
-0.7429 -0.7439 -0.7430 -0.7402 -0.7356 -0.7291 -0.7207
-0.7106 -0.6986 -0.6849 -0.6694 -0.6523 -0.6335 -0.6131
-0.5912 -0.5677 -0.5429 -0.5167 -0.4891 -0.4603 -0.4304
-0.3994 -0.3674 -0.3344 -0.3006 -0.2660 -0.2308 -0.1950
-0.1587 -0.1220 -0.0850 -0.0477 -0.0104 0.0270 0.0643
0.1014 0.1383 0.1749 0.2110 0.2465 0.2815 0.3157
0.3491 0.3817 0.4133 0.4438 0.4733 0.5015 0.5285
0.5541 0.5783 0.6011 0.6223 0.6420 0.6601 0.6765

0.6912 0.7041 0.7153 0.7247 0.7322 0.7379 0.7417
0.7436 0.7437 0.7419 0.7382
U4 =
0.7382
V4 =
-0.4592

% %Given
g=9.8;
k=125;
h0=5;
Ma=2;
Mb=5;
t1=1.009;
Va1=sqrt(2*g*h0);
Ua2=5.094;
Ub2=0;
Va2=-2.83;
Vb2=5.1;%10.61
wn=sqrt(k/Mb);
Tn=2*pi/wn;
h0=5;
t01=linspace(0,1.009,100);
h=h0-1/2*g.*t01.^2;
plot(t01,h,'m')
hold on
F=@(t)Va2.*t+(g*t.^2)/2-(Vb2./wn)*sin(wn*t);
%tI
tI=fsolve(F,0.5);
t=linspace(0,tI,100);
tt=t+1.009;
Tmeet = 1.009+tI
Fs=(Vb2./wn).*sin(wn.*t);
plot(tt,Fs,'r');
hold on
Fu=Va2.*t+(g*t.^2)/2;
plot(tt,Fu,'k');
hold on
xlabel ('time')
title ('combined motion of small and big mass')
legend ('before impact for small mass','after impact for big mass','afterimpact for smal
ler mass','location','southOutside')
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
Tmeet =
1.6186

%%%%
function hw4_prooblem2_a
g=9.81;
m=5;
L=2;
A=g/L;
xo=[pi/18,0];
[t,x]=ode45(@DE2,[0:0.01:17],xo);
plot(t,x(:,1),'r--','Linewidth',4)
to=pi/18;

td=0;
T=0:0.2:17;
th=td*sin(sqrt(A)*T)+to*cos(sqrt(A)*T);
plot(T,th,'b-','Linewidth',2)
legend('Nonlinear','Linear')
function dxdt=DE2(t,x)
g=9.81;
m=5;
L=2;
A=g/L;

dxdt=[x(2);-A*sin(x(1))];
xlabel('Time (t)'),ylabel('Response (rad)'),
grid on
title('Numerical Solution of the Pendulum for theta=10')
axis([0,18,-0.2,0.2])
hold on

function hw4_problem2_b
g=9.81;
m=5;
L=2;
A=g/L;
xo=[pi/3,0];
[t,x]=ode45(@DE2,[0:0.01:17],xo);
plot(t,x(:,1),'r--','Linewidth',2)
to=pi/3;
td=0;
T=0:0.2:17;
th=td*sin(sqrt(A).*T)+to*cos(sqrt(A)*T);
plot(T,th,'b-','Linewidth',2)
g=9.81;
m=5;
L=2;
A=g/L;
dxdt=[x(2);-A*sin(x(1))];
xlabel('Time (t)'),ylabel('Response (rad)'),
grid on
title('Numerical Solution of the Pendulum for theta=60')
axis([0,18,-1.5,1.5])
hold on

td=0;
T=0:0.2:17;
th=td*sin(sqrt(A)*T)+to*cos(sqrt(A)*T);
plot(T,th,'bo','Linewidth',2)
g=9.81;
m=5;
L=2;
A=g/L;

Time (t)
0 2 4 6 8 10 12 14 16 18
Response(rad)
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
Numerical Solution of the Pendulum for 3=10
Nonlinear
Linear

m1=2;m2=4;k1=100;k2=100;k3=200;
[frequency,v]= eigen;
M=[m1,0;0 m2];
K=[k1+k2,-k2;-k2,k2+k3];
x=[1;0];
xdot=[0;-2];
Xo=[x;xdot];
T= (2*pi/7.0711);
time=[0:0.01:3*T];
[t,q]=ode45(@mdof_ini_ode45,time,Xo);
n=length(x);
XX=q(:,(1:n));
XXdot=q(:,(1:n)+n);
q1an=0.667*cos(11.1803*t)+(1.3334/11.1803)*sin(11.1803*t)+0.3332*...
cos(7.0711*t)-(1.3334/7.0711)*sin(7.0711*t);
plot(t,q1an,'--','linewidth',2)
hold on
plot(time,q(:,1),'linewidth',1)
grid on
title('bfResponse 1-IniCond')
xlabel('bf Time period'); ylabel('bfU_1(t)')
legend('bf Analytical','bf Numerical')
hold off
q2an=-0.4998*(0.667*cos(11.1803*t)+(1.3334/11.1803)*sin(11.1803*t))...
+0.3332*cos(7.0711*t)-(1.3334/7.0711)*sin(7.0711*t);
figure
plot(t,q2an,'--','linewidth',2)
hold on
plot(time,q(:,2))
grid on
hold off
M =
2 0
0 4
K =
200 -100
-100 300
v =

-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803

m1=2;m2=4;k1=100;k2=100;k3=200;
M=[m1,0;0 m2];
K=[k1+k2,-k2;-k2,k2+k3];
M1=v.'*M*v;
K1=v.'*K*v;
x=[0;0];
xdot=[0;0];
Xo=[x;xdot];
qt=inv(v)*x;
qdot=inv(v)*xdot;
T= (2*pi/7.0711);
time=[0:0.01:3*T];
[t,q]=ode45(@mdof_step_ode45,time,Xo);
n=length(x);
XX=q(:,(1:n));
XXdot=q(:,(1:n)+n);
hold on
q1ana=0.1*(1-cos(7.0711*t));
plot(t,q1ana,'.','linewidth',2)
grid on
title('bfResponse 1-stepforce')
legend('bf Numerical','bf Analytical','location','southOutside')
hold off
figure
hold on
q2ana=0.1*(1-cos(7.0711*t));
plot(t,q2ana,'.','linewidth',2)
grid on
title('bfResponse 2-stepforce')
legend('bf Numerical','bf Analytical','location','southOutside')
hold off
M =
2 0
0 4
K =
200 -100
-100 300

v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =

7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887

D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =

2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =

50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4

K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000

frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100

-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711

frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =

-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803

M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887

2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0

0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4

K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =

7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300

v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711

D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0

0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000

frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =

200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =

7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774

-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803

M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =

200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711

frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300
v =
-0.4082 -0.5774
-0.4082 0.2887
D =
50.0000 0
0 125.0000
frequency =
7.0711
frequency =
7.0711 11.1803
M =
2 0
0 4
K =
200 -100
-100 300

m1=2;m2=4;k1=100;k2=100;k3=200;
M=[m1,0;0 m2];
K=[k1+k2,-k2;-k2,k2+k3];
x=[0;0];
xdot=[0;0];
Xo=[x;xdot];
T= (2*pi/7.0711);
time=[0:0.01:3*T];
[t,q]=ode45(@mdof_cos_ode45,time,Xo);
n=length(x);
XX=q(:,(1:n));
XXdot=q(:,(1:n)+n);
q1an=0.01185*cos(3.533*t)-0.01185*cos(11.1803*t)+0.0311*cos(3.533*t)...
-0.0311*cos(7.0711*t);
plot(t,q1an,'o','linewidth',2)
hold on
grid on
hold off
q2an=-0.4998*(0.01185*cos(3.533*t)-0.01185*cos(11.1803*t))...
+0.0311*cos(3.533*t)-0.0311*cos(7.0711*t);
figure
plot(t,q2an,'o','linewidth',2)
hold on
plot(time,q(:,2))
grid on
hold off
M =
2 0
0 4
K =
200 -100
-100 300
v =

w=linspace(0,11.1803,200);
T= (2*pi/7.0711);
t=linspace(0,3*T,200);
u1=(4./(375-3*w.^2));
u2=(7./(300-6*w.^2));
u11=u1+u2;
u22=-0.4998*u1+u2;
semilogy(w,abs(u11))
hold on
semilogy(w,abs(u22))
xlabel ('frequency responce')
ylabel ('amplitude')
title ('semilogy')
legend ('f1=5*cos(wt)','f2= 2 *cos(wt)')
grid on
hold off
figure
plot(w,u11)
xlabel ('frequency responce')
ylabel ('amplitude')
title ('linear scale')
hold on
plot(w,u22)
legend ('f1=5*cos(wt)','f2= 2 *cos(wt)')

syms r x
r=3
E=1e7;%Youngs Modulus
Rho=0.1/386.6; %Mass Density
L=100;%Length
A=2; %Area
F=100; %Force
u0=F*x/(A*E); %Intial Displacement
v0=0; %Initial Velocity
L_=pi*(r-0.5)/L; %Eigen Value of r_th Mode
W_=L*sqrt(E/Rho); %Natural Frequency of r_th Mode
Psi=vpa(sin(L_*x),5); %Eigen Function
disp ('the value of GMr')
GMr=vpa(int(Rho*A*Psi^2,0,100),5)
G=1/GMr;
disp ('the value of qr(0)')
qr_0=vpa((int(Rho*A*Psi*u0,0,100)*G),5);
pretty(qr_0)
disp ('the value of qvr(0)')
qvr_0=vpa((int(Rho*A*Psi*v0,0,100)*G),3)
r =
3
the value of GMr
GMr =
0.025867
the value of qr(0)
0.000016211
the value of qvr(0)
qvr_0 =
0.0

syms r x
L=100;%Length
A=2; %Area
F=100; %Force
L_r=pi*(r-0.5)/L; %Eigen Value of r_th Mode
W_r=L*sqrt(E/Rho); %Natural Frequency of r_th Mode
Psi=vpa(sin(L_r*x),5); %Eigen Function
disp('the value of GMr')
GMr=vpa(int(Rho*A*Psi^2,0,100),5)
Psi_a=subs(Psi,x,L);
disp('the value of GFr')
GFr=vpa(Psi_a*F,5)
the value of GMr
GMr =
0.025867 - (0.0082336*sin(6.2832*r - 3.1416))/(2.0*r - 1.0)
the value of GFr
GFr =
100.0*sin(3.1416*r - 1.5708)

clear all
clc
syms r x
L=100;%Length
A=8; %Area
F=100;%Force
a=80; b=20;% points on the line
L_r=pi*r/L; %Eigen Value of r_th Mode
M=Rho*A*L;%mass
I=M*L^2/3;%moment of inertia
Wn=(L_r*L)*sqrt(E*I/(Rho*A*L^4));
disp('the value of GMr is')
GMr=vpa(int(Rho*A*Psi^2,0,100),5)%r_th Modal Mass
u01=((F*b*x)/(6*L*E*I))*(L^2-x^2-b^2);
u02=((F*b)/(6*L*E*I))*((L/b)*(x-a)^3+(L^2-b^2)*x-x^3);
v0=0;
G=1/GMr;
qr_01=vpa(int((Rho*A*Psi*u01)*G,0,80),5);
qr_02=vpa(int((Rho*A*Psi*u02)*G,80,100),5);
disp('the value of qr(0)is')
qr_0=vpa(qr_01+qr_02,5)
disp ('the value of qvr(0) is ')
qvr_0=vpa((int(Rho*A*Psi*v0,0,100)*G),3)
the value of GMr is
GMr =
0.10347 - (0.016467*sin(6.2832*r))/r
the value of qr(0)is
qr_0 =
(1.0e-12*(sin(2.5133*r)*(1.9454e7/r^2 - 6.1596e6/r^4) - 1.0*cos(2.5133*r)*(1.6297e7/r -
1.5481e7/r^3)))/((0.016467*sin(6.2832*r))/r - 0.10347) - (1.0e-12*(sin(2.5133*r)*(1.9454
e7/r^2 - 6.1596e6/r^4) - 1.0*cos(2.5133*r)*(1.6297e7/r - 1.5481e7/r^3) - 1.0*sin(3.1416*
r)*(3.0396e7/r^2 - 6.1596e6/r^4) + cos(3.1416*r)*(3.1831e7/r - 1.9351e7/r^3)))/((0.01646
7*sin(6.2832*r))/r - 0.10347) + (9.6e-9*((1.4399e-11*(7.0369e13*sin(2.5133*r) - 1.7686e1
4*r*cos(2.5133*r)))/r^2 - (1.4399e-11*(7.0369e13*sin(3.1416*r) - 2.2107e14*r*cos(3.1416*
r)))/r^2))/((0.016467*sin(6.2832*r))/r - 0.10347) - (1.3823e-19*(7.0369e13*sin(2.5133*r)
- 1.7686e14*r*cos(2.5133*r)))/(r^2*((0.016467*sin(6.2832*r))/r - 0.10347)) - (9.4278e-4
6*(3.2667e40*sin(2.5133*r) - 3.2667e40*sin(3.1416*r) + 2.0525e40*r*cos(3.1416*r) - 1.350
5e39*r^3*cos(3.1416*r) + 6.4482e39*r^2*sin(3.1416*r)))/(r^4*((0.016467*sin(6.2832*r))/r
- 0.10347))

the value of qvr(0) is
qvr_0 =
0.0

syms r x
L=100;%Length
A=8; %Area
F=100;%Force
a=80; b=20;
L_r=pi*r/L; %Eigen Value of r_th Mode
M=Rho*A*L;
I=M*L^2/3;
Wn=(L_r*L)*sqrt(E*I/(Rho*A*L^4));
disp('the value of GMr')
GMr=vpa(int(Rho*A*Psi^2,0,100),5)%r_th Modal Mass
Psi_a=subs(Psi,x,80);
disp ('the value of GFr')
GFr=vpa(Psi_a*F,5)
the value of GMr
GMr =
0.10347 - (0.016467*sin(6.2832*r))/r
the value of GFr
GFr =
100.0*sin(2.5133*r)

%Rohit Avadhani 1001354462
%class id 1
%Use E=10^ 7 psi, Rh0=0.1/g, g=386.4 in the following problems.
%Problem 1: Ritz solution of a tapered bar with a tip mass
%Given a tapered clamped-free bar with a constant thickness 1 in., the height
%varies linearly from 4 in at the left end to 2 in at the right end. The length of the
%bar is 20 in. A concentrated mass MA is attached at the free end (MA=half of the
%bar mass). Find the first 3 natural frequencies and the associated eigenfunctions
%by Ritz method. ( Use polynomial basis functions.)Plot the eigenfunction and write
%the natural frequency in the titles of these plots. Additionally, reported the
%computed [Ka], and [Ma] and use these matrices to find an upper bound of the
%lowest natural frequency.
syms x
E=10^7;Rho=0.1/386.4;L=20;
A=4-2*x/L;I=((4-2*x/L)^3)/12;
phi0=[1 x x^2];
fBC=x;
phi=fBC*phi0;
phi_add=subs(phi,x,L);
Madd=30*Rho*(phi_add)'*(phi_add);
Ka=int((diff(phi,1))'*diff(phi,1)*E*A,'x',0,L);
Ma=int(phi'*phi*Rho*A,'x',0,L)+Madd;
[PP,EE]=eig(double(Ka),double(Ma));
[EG,ii]=sort(diag(EE));
PhiN=PP(:,ii);
PSI=phi*PhiN;
WnRitz=sqrt(EG);
for i=1:3
disp(['Eigenfunction Psi',int2str(i),'(x)= '])
vpa(PSI(i),4)
disp(' or')
vpa(expand(PSI(i)),4)
end
for i=1:length(PSI)
figure
ezplot(PSI(i),[0 L])
title(['bfEigenfunction psi',int2str(i),'(x),omega_n=',num2str(WnRitz(i))])
grid on
end
ph=PSI(1,1);
K0=int(((diff(ph))^2)*E*A,'x',0,L);
M0=int((ph^2)*Rho*A,'x',0,L)+30*Rho;

Rq=double(K0/M0);
W_upper=sqrt(Rq)
Eigenfunction Psi1(x)=
ans =
- 0.0001533*x^3 + 0.005258*x^2 + 0.412*x
or
ans =
- 0.0001533*x^3 + 0.005258*x^2 + 0.412*x
ans =
0.0009581*x^3 + 0.0977*x^2 - 2.085*x
or
ans =
0.0009581*x^3 + 0.0977*x^2 - 2.085*x
ans =
0.03051*x^3 - 0.8872*x^2 + 5.683*x
or
ans =
0.03051*x^3 - 0.8872*x^2 + 5.683*x
W_upper =
1.8499e+04

%Rohit Avadhani
%class id 1
%Problem 2: Ritz solution of a tapered bar with a tip mass, use bar eigenfunctions
%Resolve Problem 1 by using the first 3 eigenfunctions of a uniform clamped-free
%bar as basis functions.
syms x
E=10^7;Rho=0.1/386.4;L=20;
A=4-2*x/L;I=((4-2*x/L)^3)/12;
phi=[sin((pi*x)/2/L) sin((3*pi*x)/2/L) sin((5*pi*x)/2/L)];
phi_add=subs(phi,x,L);
Madd=30*Rho*(phi_add)'*(phi_add);
Ka=int((diff(phi,1))'*diff(phi,1)*E*A,'x',0,L);
Ma=int(phi'*phi*Rho*A,'x',0,L)+Madd;
PhiN=PP(:,ii);
PSI=phi*PhiN;
WnRitz=sqrt(EG);
for i=1:3
vpa(PSI(i),4)
disp(' or')
end
for i=1:length(PSI)
figure
grid on
end
ph=PSI(1,1);
K0=int(((diff(ph))^2)*E*A,'x',0,L);
M0=int(ph^2*Rho*A,'x',0,L)+30*Rho;
Rq=double(K0/M0);
W_upper=sqrt(Rq)
ans =
7.707*sin(0.07854*x) + 0.243*sin(0.3927*x) - 0.9144*sin(0.2356*x)

or
ans =
7.707*sin(0.07854*x) + 0.243*sin(0.3927*x) - 0.9144*sin(0.2356*x)
ans =
5.765*sin(0.07854*x) - 1.869*sin(0.3927*x) + 8.31*sin(0.2356*x)
or
ans =
5.765*sin(0.07854*x) - 1.869*sin(0.3927*x) + 8.31*sin(0.2356*x)
ans =
9.514*sin(0.3927*x) - 3.105*sin(0.07854*x) + 4.204*sin(0.2356*x)
or
ans =
9.514*sin(0.3927*x) - 3.105*sin(0.07854*x) + 4.204*sin(0.2356*x)
W_upper =
1.8073e+04

%Rohit Avadhani
%class id 1
%Problem 3 Ritz solution of a tapered simply-supported beam
%Given a tapered simply supported beam with a constant thickness 1 in. and height
%varies linearly from 4 in at the left end to 2 in at the right end. The length of the
%beam is 50 in. Find the first 3 natural frequencies and the associated
%eigenfunctions by Ritz method. ( Use polynomial basis functions.)Plot the
%eigenfunction and write the natural frequency in the titles of these plots.
%Additionally, reported the computed [Ka], and [Ma] and use these matrices to
%find an upper bound of the lowest natural frequency
syms x
E=10^7;Rho=0.1/386.4;L=50;
A=4-2*x/L;I=((4-2*x/L)^3)/12;
phi0=[1 x x^2];
fBC=x*(x-L);
phi=fBC*phi0;
Ka=int((diff(phi,2))'*diff(phi,2)*E*I,'x',0,L);
Ma=int(phi'*phi*Rho*A,'x',0,L);
PhiN=PP(:,ii);
PSI=phi*PhiN;
WnRitz=sqrt(EG);
for i=1:3
vpa(PSI(i),4)
disp(' or')
end
for i=1:length(PSI)
figure
grid on
end
ph=PSI(1,1);
K0=int((diff(ph,2)^2)*E*I,'x',0,L);
M0=int(ph^2*Rho*A,'x',0,L);
Rq=double(K0/M0);
W_upper=sqrt(Rq)

ans =
0.007819*x*(x - 50.0) + 0.0002051*x^2*(x - 50.0) - 2.465e-6*x^3*(x - 50.0)
or
ans =
- 2.465e-6*x^4 + 0.0003283*x^3 - 0.002436*x^2 - 0.3909*x
ans =
2.529e-5*x^3*(x - 50.0) - 7.245e-5*x^2*(x - 50.0) - 0.01556*x*(x - 50.0)
or
ans =
2.529e-5*x^4 - 0.001337*x^3 - 0.01194*x^2 + 0.778*x
ans =
0.05965*x*(x - 50.0) - 0.005403*x^2*(x - 50.0) + 0.000106*x^3*(x - 50.0)
or
ans =
0.000106*x^4 - 0.0107*x^3 + 0.3298*x^2 - 2.983*x
W_upper =
656.7699

%Rohit Avadhani
%class id -1
%Problem 4 Ritz solution of a tapered simply-supported beam use beam eigenfunctions
%Resolve Problem 3 by using the first 3 eigenfunctions of a uniform simplysupported
%beam as basis functions
syms x
E=10^7;Rho=0.1/386.4;L=50;
A=4-2*x/L;I=((4-2*x/L)^3)/12;
phi0=[sin((pi*x)/L) sin((2*pi*x)/L) sin((3*pi*x)/L)];
fBC=x*(x-L);
phi=fBC*phi0;
Ka=int((diff(phi,2))'*diff(phi,2)*E*I,'x',0,L);
Ma=int(phi'*phi*Rho*A,'x',0,L);
PhiN=PP(:,ii);
PSI=phi*PhiN;
WnRitz=sqrt(EG);
for i=1:3
vpa(PSI(i),4)
disp(' or')
end
for i=1:length(PSI)
figure
grid on
end
ph=PSI(1,1);
K0=int((diff(ph,2)^2)*E*I,'x',0,L);
M0=int(ph^2*Rho*A,'x',0,L);
Rq=double(K0/M0);
W_upper=sqrt(Rq)
ans =
0.01317*x*sin(0.06283*x)*(x - 50.0) - 0.002609*x*sin(0.1257*x)*(x - 50.0) + 0.0004586*x*
sin(0.1885*x)*(x - 50.0)

or
ans =
0.1305*x*sin(0.1257*x) - 0.6586*x*sin(0.06283*x) - 0.02293*x*sin(0.1885*x) + 0.01317*x^2
*sin(0.06283*x) - 0.002609*x^2*sin(0.1257*x) + 0.0004586*x^2*sin(0.1885*x)
ans =
0.0154*x*sin(0.1257*x)*(x - 50.0) - 7.119e-5*x*sin(0.06283*x)*(x - 50.0) - 0.003314*x*si
n(0.1885*x)*(x - 50.0)
or
ans =
0.003559*x*sin(0.06283*x) - 0.7702*x*sin(0.1257*x) + 0.1657*x*sin(0.1885*x) - 7.119e-5*x
^2*sin(0.06283*x) + 0.0154*x^2*sin(0.1257*x) - 0.003314*x^2*sin(0.1885*x)
ans =
0.004877*x*sin(0.06283*x)*(x - 50.0) + 0.001328*x*sin(0.1257*x)*(x - 50.0) + 0.01656*x*s
in(0.1885*x)*(x - 50.0)
or
ans =
0.004877*x^2*sin(0.06283*x) - 0.06642*x*sin(0.1257*x) - 0.8282*x*sin(0.1885*x) - 0.2439*
x*sin(0.06283*x) + 0.001328*x^2*sin(0.1257*x) + 0.01656*x^2*sin(0.1885*x)
W_upper =
1.4988e+03

Contents
◾ HW#6-Problem 1:Eigen value solution for 4dof spring mass system:
◾ Plot Mode shapes
◾ plotting 2nd Mode shape
◾ Mass,Stiffness and Force Normalization of Modal Matrix
clear
HW#6-Problem 1:Eigen value solution for 4dof spring mass system:
m1=10;m2=2;m3=1;m4=1;
k1=50;k2=50;k3=10;k4=10;
% Mass Matrix
M=[10,0,0,0;
0,2,0,0;
0,0,1,0;
0,0,0,1];
% K Matrix
K= [100,-50,0,0;
-50 ,70,-10,-10;
0,-10,10,0;
0,-10,0,10;];
% Initial Conditions
u0=[0;0;0;0];v0=[0;0;0;0];
disp('Initial displacement')
disp([u0])
disp('Initial velocity')
disp([v0])
% Finding Eigen Values and Eigen Vectors, use below commands
% Solution of eigenvalue problem by MATLABb function eig
[Phi1,Eeg1]=eig(K,M)
% get EigenValues (diagonal of Eigen value matrix Eeg1)
eeg=diag(Eeg1)
% Sort eigenvaluee
[Eg,Ie]=sort(eeg)
% Reorder eigenvectors acording to the order of eigenvalues
Phi=Phi1(:,Ie)
% Finding Natural Frequency
W=sqrt(Eg)
% display eigen solutions

disp('EigenSolutions: ')
disp('EigenValue Natural-frequency')
disp([Eg sqrt(Eg)])
disp('Modal matrix')
disp([Phi])
disp('Max-normalized modal matrix')
NDOF=length(Eg)
% Eigen Vectors Normalization, here Normalizing to Maximum element
% Normalize to make max element 1
for i=1:NDOF
pi=Phi(:,i);
[pm,ii]=max(abs(pi));
phis=pi*sign(pi(ii))/pm;
PhiMax(:,i)=phis;
end
PhiMax
Phi1=PhiMax(:,1); % 1st Mode shape
Phi2=PhiMax(:,2); % 2nd Mode shape
% Orthogonality Verification
%(Note: Orthogonality Verification can be done with or without
%normalization of eigen vectors)
% check M-orthogonality and K-orthogonality of the two mode shapes
disp('Check M-orthogonality and K-orthogonality of the two mode shapes')
M_ortho=Phi1'*M*Phi2
K_ortho=Phi1'*K*Phi2
Initial displacement
0
0
0
0
Initial velocity
0
0
0
0
Phi1 =
0.2134 0.2015 -0.0620 0.1000
0.2999 -0.0000 -0.0000 -0.6403
0.4269 -0.7116 -0.5209 0.1999
0.4269 -0.2959 0.8308 0.1999

Eeg1 =
2.9744 0 0 0
0 10.0000 0 0
0 0 10.0000 0
0 0 0 42.0256
eeg =
2.9744
10.0000
10.0000
42.0256
Eg =
2.9744
10.0000
10.0000
42.0256
Ie =
1
2
3
4
Phi =
0.2134 0.2015 -0.0620 0.1000
0.2999 -0.0000 -0.0000 -0.6403
0.4269 -0.7116 -0.5209 0.1999
0.4269 -0.2959 0.8308 0.1999
W =
1.7246
3.1623
3.1623
6.4827
EigenSolutions:
EigenValue Natural-frequency
2.9744 1.7246
10.0000 3.1623
10.0000 3.1623
42.0256 6.4827

Modal matrix
0.2134 0.2015 -0.0620 0.1000
0.2999 -0.0000 -0.0000 -0.6403
0.4269 -0.7116 -0.5209 0.1999
0.4269 -0.2959 0.8308 0.1999
Max-normalized modal matrix
NDOF =
4
PhiMax =
0.5000 -0.2832 -0.0746 -0.1561
0.7026 0.0000 -0.0000 1.0000
1.0000 1.0000 -0.6270 -0.3122
1.0000 0.4158 1.0000 -0.3122
Check M-orthogonality and K-orthogonality of the two mode shapes
M_ortho =
-3.8858e-16
K_ortho =
-1.1546e-14
Plot Mode shapes
plotting 1st Mode shape
figure
plot([0 1 2],[0 Phi1(1) Phi1(2)],'r--','linewidth',2)
hold on
plot(1,Phi1(1),'bo','linewidth',2)
plot(0,0,'bo','linewidth',2)
xlabel('bfDOF'),ylabel('bfMode shape coefficient')

plotting 2nd Mode shape
figure
plot([0 1 2],[0 Phi2(1) Phi2(2)],'r--','linewidth',2)
hold on
plot(0,0,'bo','linewidth',2)
xlabel('bfDOF'),ylabel('bfMode shape coefficient')

Mass,Stiffness and Force Normalization of Modal Matrix
disp('Mass-normalization of modal matrix')
GM=Phi.'*M*Phi
disp('Stiffness-normalization of modal matrix')
GK=Phi.'*K*Phi
% Initial Modal Displacement and velocity
disp('Modal displacement')
q0=Phi'*M*u0
disp('Modal velocity')
qv0=Phi'*M*v0
% Part C
Z=K-10*M
%syms v v1 v2 v3 v4
%v=[v1,0,0,0;0,v2,0,0;0,0,v3,0;0,0,0,v4]
%Z*v={0}
P2=[1;0;1;-6]
P3=[1;0;-6;1]
P2.'*M*P3

Mass-normalization of modal matrix
GM =
1.0000 0.0000 -0.0000 0.0000
0.0000 1.0000 -0.0000 -0.0000
-0.0000 -0.0000 1.0000 -0.0000
0.0000 -0.0000 -0.0000 1.0000
Stiffness-normalization of modal matrix
GK =
2.9744 0.0000 -0.0000 0.0000
0.0000 10.0000 -0.0000 0.0000
-0.0000 -0.0000 10.0000 -0.0000
0.0000 0.0000 -0.0000 42.0256
Modal displacement
q0 =
0
0
0
0
Modal velocity
qv0 =
0
0
0
0
Z =
0 -50 0 0
-50 50 -10 -10
0 -10 0 0
0 -10 0 0
P2 =
1
0
1
-6

P3 =
1
0
-6
1
ans =
-2

Contents
◾ HW#6-Problem 2:Eigen value solution for 4dof spring mass system:
%%Rohit Avadhani 1001354462
HW#6-Problem 2:Eigen value solution for 4dof spring mass system:
m1=10;m2=2;m3=1;m4=1;
k1=50;k2=50;k3=10;k4=10;
% Mass Matrix
M=[10,0,0,0;
0,2,0,0;
0,0,1,0;
0,0,0,1];
% K Matrix
K= [100,-50,0,0;
-50 ,70,-10,-10;
0,-10,10,0;
0,-10,0,10;];
% Initial Conditions
u0=[0;0];v0=[0;0];
disp('Initial displacement')
disp([u0]);
disp('Initial velocity')
disp([v0]);
% Finding Eigen Values and Eigen Vectors, use below commands
% Solution of eigenvalue problem by MATLABb function eig
[Phi1,Eeg1]=eig(K,M);
% get EigenValues (diagonal of Eigen value matrix Eeg1)
eeg=diag(Eeg1);
% Sort eigenvaluee
[Eg,Ie]=sort(eeg);
% Reorder eigenvectors acording to the order of eigenvalues
Phi=Phi1(:,Ie) ;
% Finding Natural Frequency
W=sqrt(Eg)
% display eigen solutions
disp('EigenSolutions: ');
disp('EigenValue Natural-frequency');
disp([Eg sqrt(Eg)]);
disp('Modal matrix');
disp([Phi]);

disp('Max-normalized modal matrix');
NDOF=length(Eg);
% Re-Normalizing Eigen vector:
i=1;
while i<=4
E=ones(4,1);
aa=Phi(:,i)'*M*E;
PhiR(:,i)=aa*Phi(:,i)
GMRi=PhiR(:,i)'*M*PhiR(:,i)
i=i+1;
end
Initial displacement
0
0
Initial velocity
0
0
W =
1.7246
3.1623
3.1623
6.4827
EigenSolutions:
EigenValue Natural-frequency
2.9744 1.7246
10.0000 3.1623
10.0000 3.1623
42.0256 6.4827
Modal matrix
0.2134 0.2015 -0.0620 0.1000
0.2999 -0.0000 -0.0000 -0.6403
0.4269 -0.7116 -0.5209 0.1999
0.4269 -0.2959 0.8308 0.1999
Max-normalized modal matrix
PhiR =
0.7659
1.0762
1.5318
1.5318

GMRi =
12.8747
PhiR =
0.7659 0.2030
1.0762 -0.0000
1.5318 -0.7170
1.5318 -0.2981
GMRi =
1.0151
PhiR =
0.7659 0.2030 0.0192
1.0762 -0.0000 0.0000
1.5318 -0.7170 0.1614
1.5318 -0.2981 -0.2574
GMRi =
0.0960
PhiR =
0.7659 0.2030 0.0192 0.0119
1.0762 -0.0000 0.0000 -0.0762
1.5318 -0.7170 0.1614 0.0238
1.5318 -0.2981 -0.2574 0.0238
GMRi =
0.0141

Contents
◾ Four DOF_Data_1_IC.m
◾ Data for 4-DOF system:
◾ system data
◾ Initial conditions
◾ Forcing functions
◾ Time duration for pulse ( non-zero force from 0 to T0)
◾ time points
◾ Frequency points
◾ Type of analysis
◾ Output specification
◾ Output specification
◾ -------------
Four DOF_Data_1_IC.m
Data for 4-DOF system:
system data
Mass matrix
M=[10 0 0 0;
0 2 0 0;
0 0 1 0;
0 0 0 1];
% stiffness matrix
K=[ 100 -50 0 0;
-50 70 -10 -10
0 -10 10 0
0 -10 0 10];
% Damping matrix
C=[2 0 0 0;0 1 0 -1;0 0 0 0;0 -1 0 1]
C =
2 0 0 0
0 1 0 -1
0 0 0 0
0 -1 0 1

Initial conditions
u0=[.1;.1;.1;.1];v0=[1;0;0;0];
Forcing functions
Static
P0=[10;0;0;2];
% Sine component
Ps=[0;0;1;0]; % amplitude
Ws=4; % frequency, rad/s
% Cosine component
Pc=[10;0;0;2]; % amplitude
Wc=4; % frequency, rad/s
Time duration for pulse ( non-zero force from 0 to T0)
T0=100;
time points
tALL=linspace(0,10,200);
Frequency points
WALL=0;
Type of analysis
Analysis='Analytical'; % for time domain solution
% Analysis='Numerical';
% Analysis='Symbolic';
% Analysis='Frequency Response'; % for frequency domain solution
% Analysis='Pulse Response'; % for pulse inputs
% Analysis='Modal analysis'; % for eigenvalue problem
% Analysis='Spring-mass-damper'; % forulation of M,K,C matrices
Output specification
PrintDOF=[0 ]; % ID of outpit dof
% Print time domain solution of PrintDOF at time tALL
% Or
% Print frequency domain solution of PrintDOF at frequency WALL time tn

Output specification
PlotDOF=[1;2;3;4 ];
% ID of outpit dof
% Plot time domain solution of PlotDOF at time tALL
% Or
% Plot frequency domain solution of PlotDOF at frequency WALL time tn
-------------

syms r x
E=1e7;%Youngs modulus
Rho=0.1/386.6;%density
L=100;%Length
A=2;%cross sectional area
L_=pi*(r-0.5)/L;% eigen values
W_=L*sqrt(E/Rho);% natural frequencies
psi=sin(L_*x);%eigen functions
Psi=vpa(psi,5);% numerical values of eigen functions
% plotting for third function
disp('the plot of third eigen function is as follows')
r=3;%Third mode value
x=linspace(0,L,100);
Psi_=sin(0.031416.*x.*(r - 0.5));%eigen function vector
plot(x,Psi_,'k*');
grid on
xlabel('bfL')
ylabel('bfPsi(x)')
title ('Eigen function to Length')
the plot of third eigen function is as follows

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