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CVEN9840-Lacture4-2020copy.pdf
Structural Health Monitoring
(CVEN9840)
Lecture 4
Dr Mehri Makki Alamdari
School of Civil and Environmental Engineering
1
2CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Structural Dynamics
Multi Degree of Freedom (MDOF)
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Multi Degree of Freedom System (MDOF)
¾ In the previous lecture we used SDOF system to describe time response of a vibrating system. However, SDOF is only applicable if movement of a structure or a system can be modelled with only one single coordinate.
¾ However, if the deflected shape of the structure involves more than one single coordinate, a SDOF system will only provide an approximate solution and will not be a true dynamic representative of the structure.
Dynamic analysis of most real structures is based on multiple-degree-of- freedom (MDOF) models.
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[K] is the stiffness matrix
[C] is the damping matrix
[M] is the mass matrix
Multi Degree of Freedom System (MDOF)
Finite element approach is often adopted to establish mass, stiffness and damping matrices.
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A three-story building shown below is modelled as a 3 Degree-of-Freedom (DOF) system. Derive equation of motion for this system.
Example
P2P1 P3
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Solution 1. Draw a free-body diagram of each mass
3. For the three springs, relate the linearly elastic spring force to the displacement.
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Solution
4. Combine the equations.
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Example
A two-story building as seen in figure below is subjected to a ground motion of z as a res l of an ear hq ake. Use Ne on s La s o deri e he eq a ion of mo ion for he system.
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Solution
1. Let the relative displacements, that is the displacements of the two masses relative to the moving ground, be defined by:
2. For each mass, draw a free-body-diagram.
3. Use Ne on s Second la o ri e he eq a ions of mo ion.
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Solution
4. Combine and simplify the equations:
5. Write the above equations in matrix form:
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Natural Frequencies and Mode Shapes
¾ Undamped MDOF system
Assume the solution to the equation of motion is in the form of simple harmonic function:
Here {𝜙 is the mode shape vector describing the relative movement of different degrees of freedom with respect to each other.
known as an eigenvalue problem
The trivial solution
The non-trivial solution
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¾ There will be N roots, i.e. ω2 (i = 1, 2, , N ) hich are called he eigen al es.
¾ The ω is called the i-th eigenfrequency which is the natural frequency of the vibration.
¾ For each natural frequency ω , there is a corresponding eigenvector ϕ which is called the mode shape and can be determined from the following equation:
Natural Frequencies and Mode Shapes
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Natural Frequencies and Mode Shapes
If 𝜙 is a solution, 𝛼 𝜙 is also a solution.
� An eigenvector can be normalized by any number.
� If all the eigenvalues are arranged in the matrix form, this resulting matrix is called the spectral matrix and is a diagonal matrix.
� The corresponding eigenvectors arranged in the matrix form is called the modal matrix and is in the form of a square matrix.
� The relationship between the modal and the spectral matrices.
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Obtain the natural frequencies and mode shapes of the 2-DOF system shown below.
k1=k2=k3=k m1=m2=m
For a particular case of k=1,000N/m and m=10kg obtain the results.
Example
u1(t) u2(t)
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Solution
¾ The equation of motion in matrix form will be:
¾ Assume a harmonic motion:
Note: This means that when the 2-dof system vibrates at natural frequency ω, u1and u2 have the same time dependence and at all times the amplitude ratio of u2 and u1 has the same value.
¾ Substitute in the equation of motion:
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Solution
¾ Set up the characteristic equation, which is the determinant of the coefficient matrix:
¾ Expand the determinant to get the characteristic equation:
¾ Obtain the roots of the characteristic equation:
¾ Obtain mode shapes:
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Solution
¾Notice that the original system is symmetric about the centre of the middle spring. For this reason the mode shapes turn out to be a symmetric mode (mode 1) and an antisymmetric mode (mode 2). This is an important result, since many structures possess such physical symmetry. For example, an airplane is symmetric (nominally) about a vertical plane through the axis of the fuselage.
¾Also note that mode 2 has a node point, that is, a point that always remains stationary.
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Standard eigenvalue problem in MATLAB Æ [phi,lambda]=eig(K,M);
lambda is a diagonal matrix containing eigenvalues Æ phi is a square matrix containing corresponding eigenvectors in each column
Solution in MATLAB
clc; clear all; close all; k=1000; %N/m m=10; %kg K=[2*k -k; -k 2*k]; M=[m 0; 0 m]; [phi,lambda]=eig(K,M); w1=lambda(1,1).^0.5 %rad/sec w2=lambda(2,2).^0.5 %rad/sec f1=w1/(2*pi) %Hertz f2=w2/(2*pi) %Hertz phi1=phi(:,1); %first mode shape phi2=phi(:,2); %second mode shape phi1=phi(:,1)/phi(1,1) phi2=phi(:,2)/phi(1,2)
phi1_n=[0 phi1' 0]; phi2_n=[0 phi2' 0];
19CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
figure hold on plot(phi1_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,1.1,'linewidth',2,'color','red') hold off axis tight box on title ('Mode 1') set(gca,'fontsize',14) figure hold on plot(phi2_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,-1.1,'linewidth',2,'color','red') hold off set(gca, 'XAxisLocation', 'origin', 'YAxisLocation', 'origin') box on title ('Mode 2') set(gca,'fontsize',14)
20CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
w1 = 10.0000 rad/s
w2 = 17.3205 rad/s
f1 = 1.5915 Hz
f2 = 2.7566 Hz
phi1 = 1 1
phi2 = 1
-1
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Repeat the same exercise before for 3-dof and 4 dof mass-spring systems shown below. Present the mode shapes and natural frequencies if k=1000N/m and m=10kg. Assume all the springs have the same stiffness value and all masses are equal.
Example
Illustration of a 3-dof system
Illustration of a 4-dof system
k1=k2=k3=k4=k5=k m1=m2=m3=m4=m
(a)
(b)
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Solution in MATLAB
¾(a) A 3-dof system clc; clear all; close all; %3 dof system k=1000; %N/m m=10; %kg K=[2*k -k 0; -k 2*k -k ; 0 -k 2*k]; M=[m 0 0; 0 m 0; 0 0 m]; [phi,lambda]=eig(K,M); w1=lambda(1,1).^0.5 %rad/sec w2=lambda(2,2).^0.5 %rad/sec w3=lambda(3,3).^0.5 %rad/sec f1=w1/(2*pi) %Hertz f2=w2/(2*pi) %Hertz f3=w3/(2*pi) %Hertz phi1=phi(:,1); %first mode shape phi2=phi(:,2); %second mode shape phi3=phi(:,3); %third mode shape phi1=phi(:,1)/phi(1,1) phi2=phi(:,2)/phi(1,2) phi3=phi(:,3)/phi(1,3) phi1_n=[0 phi1' 0]; phi2_n=[0 phi2' 0]; phi3_n=[0 phi3' 0];
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% Plot Mode 1 figure hold on plot(phi1_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,1.5,'linewidth',2,'color','red') hold off hold on quiver(4,0,0,1.1,'linewidth',2,'color','red') hold off axis tight box on title ('Mode 1') set(gca,'fontsize',14)
Solution in MATLAB
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% Plot Mode 2 figure hold on plot(phi2_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(4,0,0,-1.1,'linewidth',2,'color','red') hold off set(gca, 'XAxisLocation', 'origin', 'YAxisLocation', 'origin') box on title ('Mode 2') set(gca,'fontsize',14)
Solution in MATLAB
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% Plot Mode 3 figure hold on plot(phi3_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,-1.4,'linewidth',2,'color','red') hold off hold on quiver(4,0,0,1.1,'linewidth',2,'color','red') hold off set(gca, 'XAxisLocation', 'origin', 'YAxisLocation', 'origin') box on title ('Mode 3') set(gca,'fontsize',14)
Solution in MATLAB
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Solution in MATLAB
phi1 =
1.0000 1.4142 1.0000
phi2 =
1.0000 0.0000 -1.0000
phi3 =
1.0000 -1.4142 1.0000
w1 = 7.6537 rad/s
w2 = 14.1421 rad/s
w3 = 18.4776 rad/s
f1 = 1.2181 Hz
f2 = 2.2508 Hz
f3 = 2.9408 Hz
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Solution in MATLAB
f1 =1.2181 Hz f2 =2.2508 Hz f3 =2.9408 Hz
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Solution in MATLAB
¾(b) A 4-dof system %4 dof system k=1000; %N/m m=10; %kg K=[2*k -k 0 0; -k 2*k -k 0 ; 0 -k 2*k -k; 0 0 -k 2*k]; M=[m 0 0 0; 0 m 0 0; 0 0 m 0; 0 0 0 m]; [phi,lambda]=eig(K,M); w1=lambda(1,1).^0.5 %rad/sec w2=lambda(2,2).^0.5 %rad/sec w3=lambda(3,3).^0.5 %rad/sec w4=lambda(4,4).^0.5 %rad/sec f1=w1/(2*pi) %Hertz f2=w2/(2*pi) %Hertz f3=w3/(2*pi) %Hertz f4=w4/(2*pi) %Hertz phi1=phi(:,1); %first mode shape phi2=phi(:,2); %second mode shape phi3=phi(:,3); %third mode shape phi4=phi(:,4); %fourth mode shape phi1=phi(:,1)/phi(1,1) phi2=phi(:,2)/phi(1,2) phi3=phi(:,3)/phi(1,3) phi4=phi(:,4)/phi(1,4) phi1_n=[0 phi1' 0]; phi2_n=[0 phi2' 0]; phi3_n=[0 phi3' 0]; phi4_n=[0 phi4' 0];
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Solution in MATLAB
% Plot Mode 1 figure hold on plot(phi1_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,1.7,'linewidth',2,'color','red') hold off hold on quiver(4,0,0,1.7,'linewidth',2,'color','red') hold off hold on quiver(5,0,0,1.1,'linewidth',2,'color','red') hold off axis tight box on title ('Mode 1') set(gca,'fontsize',14)
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Solution in MATLAB
% Plot Mode 2 figure hold on plot(phi2_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,0.7,'linewidth',2,'color','red') hold off hold on quiver(4,0,0,-0.7,'linewidth',2,'color','red') hold off set(gca, 'XAxisLocation', 'origin', 'YAxisLocation', 'origin') hold on quiver(5,0,0,-1.1,'linewidth',2,'color','red') hold off box on title ('Mode 2') set(gca,'fontsize',14)
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Solution in MATLAB
% Plot Mode 3 figure hold on plot(phi3_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,-0.7,'linewidth',2,'color','red') hold off hold on quiver(4,0,0,-0.7,'linewidth',2,'color','red') hold off hold on quiver(5,0,0,1.1,'linewidth',2,'color','red') hold off set(gca, 'XAxisLocation', 'origin', 'YAxisLocation', 'origin') box on title ('Mode 3') set(gca,'fontsize',14)
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Solution in MATLAB
% Plot Mode 4 figure hold on plot(phi4_n,'linewidth',2) hold off hold on quiver(2,0,0,1.1,'linewidth',2,'color','red') hold off hold on quiver(3,0,0,-1.7,'linewidth',2,'color','red') hold off hold on quiver(4,0,0,1.7,'linewidth',2,'color','red') hold off hold on quiver(5,0,0,-1.1,'linewidth',2,'color','red') hold off set(gca, 'XAxisLocation', 'origin', 'YAxisLocation', 'origin') box on title ('Mode 4') set(gca,'fontsize',14)
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Solution in MATLAB
f1 = 0.9836 Hz f2 = 1.8710 Hz f3 = 2.5752 Hz f4 = 3.0273 Hz
w1 = 6.1803 rad/s w2 = 11.7557 rad/s w3 = 16.1803 rad/s w4 = 19.0211 rad/s
phi1 = 1.0000 1.6180 1.6180 1.0000
phi2 = 1.0000 0.6180 -0.6180 -1.0000
phi3 = 1.0000 -0.6180 -0.6180 1.0000
phi4 = 1.0000 -1.6180 1.6180 -1.0000
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Solution in MATLAB
f4 = 3.0273 Hz
f1 = 0.9836 Hz f2 = 1.8710 Hz
f3 = 2.5752 Hz
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Understanding the Mode Shapes
Ref: https://www.youtube.com/watch?v=kvG7OrjBirI
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Mode Shapes Æ Cable-Stayed Bridge
¾ A Cable-stayed bridge
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Example
A simplified model of an automobile suspension system is shown below. The vehicle weight is 3000lb and has a centroidal moment of inertial of 300 slug.ft2 and is connected to massless springs with stiffness of 2000lb/ft. Distance between the front axle and centre of gravity (G) is 4.5ft and distance between the rear axle and centre of gravity is 3.5ft. Determine natural frequencies and mode shapes of this 2-dof system.
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Solution in MATLAB
Let x be the displacement of mass centre and 𝜃 be the angular rotation of the bar from its horizontal line. It can be shown that the equation of motion will be:
clc; clear all; close all; g=32.17; %gravity m=3000/g; I=300; b=4.5; a=3.5; k=2000; M=[m 0 ; 0 I]; %Mass matrix K=[2*k (b-a)*k; (b-a)*k (b^2+a^2)*k]; %Stiffness matrix [phi,lambda]=eig(K,M); w=lambda.^0.5; % rad/sec w1=w(1,1) %first natural frequency w2=w(2,2) %second natural frequency phi1=phi(:,1)/phi(1,1) %first mode shape phi2=phi(:,2)/phi(1,2) %second mode shape
w1 = 6.4865 rad/sec w2 = 14.7474 rad/sec
phi1 = 1.0000 -0.0382
phi2 = 1.0000 8.1408
A pitching mode
A bouncing mode
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Orthogonality of Mode Shapes
Consider two mode shapes (eigenvectors) of ϕ and ϕ corresponding to two eigenvalues 𝜔n 𝜔 m from the eigenvalue problem.
They are equal with their transpose.
Subtract
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¾ Mode shapes are orthogonal with respect to the mass and the stiffness matrices.
¾ Mode shapes are physically distinct.
¾ Mode shapes (Eigenvectors) are linearly independent Æ We cannot construct any one mode with any linear combination of the other modes.
jimM
jiM
ii T
i
j T
i
==
=
]][[][
0]][[][
jikK
jiK
ii T
i
j T
i
==
=
]][[][
0]][[][
𝜔2 𝑘 𝑚 }]{[}{
}]{[}{ 2
i T
i
i T
i i M
K =
Orthogonality of Mode Shapes
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Mass Normalised Mode Shapes
Æ Mass Normalised Mode Shapes
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Example
For the 4 dof mass-spring system solved before evaluate the orthogonality relationships.
Illustration of a 4-dof system
k1=k2=k3=k4=k5=k= 1,000N/m m1=m2=m3=m4=m= =10kg
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Solution in MATLAB for i=1:4;
for j=1:4; orthogonality_check_M(i,j)=phi(:,i)'*M*phi(:,j); orthogonality_check_K(i,j)=phi(:,i)'*K*phi(:,j);
end; end for i=1:4;
w(i)=(orthogonality_check_K(i,i)./orthogonality_check_M(i,i))^0.5; end;
orthogonality_check_M orthogonality_check_K w
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Modal Expansion
=
= N
i iiN qu
1 1 }{}{
We can expend any arbitrary vector by superposition of the eigenvectors as:
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Modal Expansion
¾ Displacement vector {u (t )} is the sum of all of the modes where 𝑦 𝑡 is a scalar multiplier called modal coordinates.
or
To determine the modal coordinates:
Orthogonality
The modal expansion of the displacement vector is used to obtain the free vibration response of an undamped system.
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Example
A uniform beam is modelled by three lumped masses on a massless beam. The model is symmetric about its central mass. The mutually orthogonal normal modes can be identified as two symmetric modes and one antisymmetric mode, as shown below. Show that the expansion of any arbitrary symmetric deflection v will involve only the symmetric modes of the structure.
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Solution
¾ We need to show that for an arbitrary symmetric displacement vector of the form:
The coefficient c2 of the antisymmetric mode vanishes in the expansion:
The coefficient c2 will be:
¾ Therefore, c2=0, and the expansion of any symmetric deflection only involves the two symmetric modes of the system.
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Undamped Free Vibration Response
Initial conditions (IC)
Modal expansion
Ak and Bk are nkno n cons an s ha can be de ermined from he IC s.
Substituting t = 0 Modal expansion
Similarly Æ
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Determine expression for the free vibration response of the 2-DOF system shown below if the initial conditions are:
Assume: k1=k2=k3=k m1=m2=m
For a particular case of k=1,000N/m and m=10kg plot the time response for 5 seconds for each degree of freedom if u0= 0.05 m.
Example
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Solution
¾ From the previous solution, the mass matrix and mode shapes for the system are:
¾ Determine the modal masses Mi
¾ Determine the coefficients Ai and Bi:
B1=B2=0
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Solution
¾ Final solution will be:
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Solution in MATLAB
clc; clear all; close all; k=1000; %N/m m=10; %kg K=[2*k -k; -k 2*k]; %stiffness matrix M=[m 0; 0 m]; %mass matrix [phi,lambda]=eig(K,M); u0=[0;0.05]; %Initial condition udot0=[0;0]; %Initial condition for i=1:2;
phi(:,i)=phi(:,i)/phi(1,i); %mode shape w(i)=lambda(i,i)^0.5; %natural frequency m(i)=phi(:,i)'*M*phi(:,i); %modal mass A(i)=phi(:,i)'*M*u0/m(i); B(i)=phi(:,i)'*M*udot0/(m(i)*w(i));
end; k=0; for t=0:0.001:5;
k=k+1; s=0;
for i=1:2; s=A(i)*phi(:,i)*cos(w(i).*t)+s;
end;
53CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
u1(k)=s(1); %1st dof response u2(k)=s(2); %2nd dof response end; t=0:0.001:5; figure subplot(211) plot(t,u1,'linewidth',2) xlabel('time sec') ylabel('u_1') set (gca,'fontsize',14) subplot(212) plot(t,u2,'linewidth',2) xlabel('time sec') ylabel('u_2') set (gca,'fontsize',14)
54CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Assignment
Determine the free vibration of an automobile shown using the pitch (angular motion) and bounce (up and down linear motion) as the DOFs. Given the mass m of the automobile to be 1500 kg, the radius of gyration r= 1.1 m, the distance between the C.G. to the front axle l1 = 1.5 m, the distance between the C.G. to the rear axle l2 = 1.6 m, the front spring stiffness kf =36 kN/m, the rear spring stiffness kr =39 kN/m. The rotational inertial force about its C.G. can be determined from 𝐼0𝜃 𝑚𝑟2 𝜃. The initial conditions are 𝑢0= 0.01 m; 𝜃0 = 0; 𝑢0 0; 𝜃0= 0.001 rad/sec. Plot the time response for each dof for 10 seconds.
55CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB clc; clear all; close all; m=1500; r=1.1; %radius of gyration l1=1.5; %distance from c.g. to front axle l2=1.6; %distance from c.g. to rear axle kf=36e3; %front axle stiffness kr=39e3; %rear axle stiffness u0=[0.01;0]; %initial displacement udot0=[0 ; 0.001]; %initial velocity M=[m 0; 0 m*r^2]; %Mass matrix K=[kf+kr -kf*l1+kr*l2;-kf*l1+kr*l2 kf*l1^2+kr*l2^2]; %Stiffness matrix [phi,lambda]=eig(K,M);
for i=1:2; phi(:,i)=phi(:,i)/phi(1,i); %mode shape w(i)=lambda(i,i)^0.5; %natural frequency m(i)=phi(:,i)'*M*phi(:,i); %modal mass A(i)=phi(:,i)'*M*u0/m(i); B(i)=phi(:,i)'*M*udot0/(m(i)*w(i));
end; k=0; for t=0:0.001:10;
k=k+1; s=0;
for i=1:2; s=A(i)*phi(:,i)*cos(w(i).*t)+s;
end; u1(k)=s(1); u2(k)=s(2); end; t=0:0.001:10;
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Solution in MATLAB
figure hold on plot(t,u1,'linewidth',2) hold off xlabel('time sec') set (gca,'fontsize',14) hold on plot(t,u2,'linewidth',2) hold off xlabel('time sec') set (gca,'fontsize',14) legend ('Bounce Motion','Pitch Motion') box on grid on grid minor
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Modal Expansion Æ Uncouple Equations of Motion
DisplacementApplied Force
¾ An example to show why modal expansion is important to solve MDOF systems:
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Coupled Equations of Motion
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Modal Expansion Æ Uncoupled Equations of Motion
Using Orthogonality Conditions Æ Uncoupled Equations of Motion
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Damped Free Vibration
Assuming the solution is in the form of:
Or
The ICs are given as:
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Damped Free Vibration
is a diagonal matrix from the orthogonality relation
is a diagonal matrix from the orthogonality relation
may or may not be a diagonal matrix
If [Cn ] is a diagonal matrix (classical viscous damping matrix ) Æ there will be N uncoupled differential equations Æ N independent SDOF systems.
If [Cn ] is not a diagonal matrix Æ there exists no analytical solution and the problem has to be solved numerically.
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¾ Consider the case of classical viscous damping:
Damped Free Vibration Æ Classical Damping
Damping ratio for the n-th mode:
From Lecture 3:
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Damped Free Vibration Æ Classical Damping
Similar to a SDOF system, the amplitude of the n-th mode free vibration of a MDOF system with classical damping will decay with the rate depending on the n-th m de damping ratio.
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Example
Obtain a free vibration solution for a four DOF system with mass and stiffness matrix shown. Assume 5% damping for each vibration mode. Plot the time response of each dof for 100 seconds.
m
The initial conditions are:
65CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB M=[4 0 0 0; 0 4 0 0; 0 0 4 0; 0 0 0 4]; %Mass matrix K=[10 -5 0 0; -5 10 -5 0;0 -5 10 -5;0 0 -5 5]; %Stiffness matrix [phi,lambda]=eig(K,M); u0=[0.025;0.02;0.01;0.001]; %Initial displacements udot0=[0;0;0;0]; %Initial velocity zeta(1)=0.05; %first mode damping factor zeta(2)=0.05; %second mode damping factor zeta(3)=0.05; %third mode damping factor zeta(4)=0.05; %fourth mode damping factor for i=1:4;
wn(i)=lambda(i,i)^0.5; %natural frequencies in rad/sec f(i)=wn(i)/(2*pi); %natural frequencies in Hz wd(i)=wn(i)*(1-zeta(i)^2)^0.5; %damped natural frequencies in rad/sec m(i)=phi(:,i)'*M*phi(:,i); %modal mass y0(i)=phi(:,i)'*M*u0/m(i); ydot0(i)=phi(:,i)'*M*udot0/m(i);
end; k=0; for t=0:0.0001:100;
k=k+1; s=0;
for i=1:4; s=phi(:,i)*exp(-zeta(i)*wn(i)*t)*(y0(i)*cos(wd(i)*t)+ (1/wd(i))*(ydot0(i)+zeta(i)*wn(i)*y0(i))*sin(wd(i)*t))+s;
end; u1(k)=s(1); %first dof u2(k)=s(2); %second dof u3(k)=s(3); %third dof u4(k)=s(4); %fourth dof end;
66CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
t=0:0.0001:100; figure subplot(411) plot(t,u1,'linewidth',2) xlabel('time sec') ylabel('u_1') set (gca,'fontsize',14) subplot(412) plot(t,u2,'linewidth',2) xlabel('time sec') ylabel('u_2') set (gca,'fontsize',14) subplot(413) plot(t,u3,'linewidth',2) xlabel('time sec') ylabel('u_3') set (gca,'fontsize',14) subplot(414)
plot(t,u4,'linewidth',2) xlabel('time sec') ylabel('u_4') set (gca,'fontsize',14) %check at the first time instant the displacements match with the initial displacement given u1(1) u2(1) u3(1) u4(1)
67CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
68CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
69CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Example
Obtain a free vibration solution for a four DOF system considering the contribution from the first two modes only. Assume 5% damping in the first two modes. Plot the time response of each dof for 100 seconds.
m
The initial conditions are:
70CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB clc; clear all; close all; M=[4 0 0 0; 0 4 0 0; 0 0 4 0; 0 0 0 4]; %Mass matrix K=[10 -5 0 0; -5 10 -5 0;0 -5 10 -5;0 0 -5 5]; %Stiffness matrix [phi,lambda]=eig(K,M); u0=[0.025;0.02;0.01;0.001]; %Initial displacement udot0=[0;0;0;0]; %initial velocity zeta(1)=0.05; zeta(2)=0.05; for i=1:2;
wn(i)=lambda(i,i)^0.5; f(i)=wn(i)/(2*pi); wd(i)=wn(i)*(1-zeta(i)^2)^0.5; m(i)=phi(:,i)'*M*phi(:,i); y0(i)=phi(:,i)'*M*u0/m(i); ydot0(i)=phi(:,i)'*M*udot0/m(i);
end; k=0; for t=0:0.0001:100;
k=k+1; s=0;
for i=1:2; s=phi(:,i)*exp(-zeta(i)*wn(i)*t)*(y0(i)*cos(wd(i)*t)+ (1/wd(i))*(ydot0(i)+zeta(i)*wn(i)*y0(i))*sin(wd(i)*t))+s;
end; u1(k)=s(1); u2(k)=s(2); u3(k)=s(3); u4(k)=s(4); end;
71CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
t=0:0.0001:100; figure subplot(411) plot(t,u1,'linewidth',2) xlabel('time sec') ylabel('u_1') set (gca,'fontsize',18) subplot(412) plot(t,u2,'linewidth',2) xlabel('time sec') ylabel('u_2') set (gca,'fontsize',18) subplot(413) plot(t,u3,'linewidth',2) xlabel('time sec') ylabel('u_3') set (gca,'fontsize',18) subplot(414)
plot(t,u4,'linewidth',2) xlabel('time sec') ylabel('u_4') set (gca,'fontsize',18) u1(1) u2(1) u3(1) u4(1) %They do not 100% match, because we are not considering the contribution from all four vibration modes!
72CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
73CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Rayleigh Proportional Damping
a0 and a1 are constants with units of sec-1 and sec.
¾ Rayleigh damping is mass-proportional and stiffness-proportional.
If the i-th and the j-th modal damping ratios and the corresponding natural frequencies are known, a0 and a1 can be obtained.
Rayleigh damping is a special case of classic viscous damping.
74CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Rayleigh Proportional Damping
75CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Example
For a four DOF system shown below, the damping ratios for the first mode and the fourth mode are, respectively, 2% and 1%. Determine the proportional Rayleigh damping matrix and calculate the damping ratio for the second and third modes.
76CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
clc; clear all; close all; M=[5 0 0 0; 0 5 0 0; 0 0 5 0; 0 0 0 5]; %Mass matrix K=[30 -7 0 0; -7 20 -10 0; 0 -10 10 -5; 0 0 -5 15]; %Stiffness matrix [phi,lambda]=eig(K,M); zeta(1)=0.02; %damping ratio for mode 1 zeta(4)=0.01; %damping ratio for mode 4 for i=1:4;
wn(i)=lambda(i,i)^0.5; %natural frequencies end; a=(0.5*[1/wn(1) wn(1); 1/wn(4) wn(4)])^-1*[zeta(1);zeta(4)]; a0=a(1); %Mass matrix coefficient in Rayleigh damping a1=a(2); %Stiffness matrix coefficient in Rayleigh damping zeta(2)=a0/(2*wn(2))+a1/2*wn(2) %damping ratio for mode 2 zeta(3)=a0/(2*wn(3))+a1/2*wn(3) %damping ratio for mode 3
C=a0*M+a1*K
77CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
78
Example
Determine the free vibration of a three-storey shear building shown. Use the Rayleigh damping matrix assuming damping ratio of 1% for the first and the third mode. The initial conditions are {𝑢(0)}= 10 12 17 cm and {𝑢(0)} = {0} .
For a particular case of k=30,000N/m and m=45,000kg plot the time response for 5 seconds for each degree of freedom.
𝑢1
𝑢2
𝑢3
CVEN9840 Structural Health Monitoring 2020 Dr Mehri Makki Alamdari
79CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB u0=0.01*[10 ;12 ;17]; %Initial displacement in m udot0=[0;0;0]; %Initial velocity k=30000; %N/m m=45000; %Kg k1=k/3; k2=7*k/9; k3=k; M=[m/2 0 0; 0 m 0; 0 0 m]; %Mass matrix K=[k1 -k1 0; -k1 k1+k2 -k2; 0 -k2 k2+k3]; %Stiffness matrix [phi,lambda]=eig(K,M); zeta(1)=0.1; %damping ratio for mode 1 zeta(3)=0.1; %damping ratio for mode 3 for i=1:3;
wn(i)=lambda(i,i)^0.5; %natural frequencies in rad/sec f(i)=wn(i)/(2*pi); %natural frequencies in Hz
end; a=(0.5*[1/wn(1) wn(1); 1/wn(3) wn(3)])^-1*[zeta(1);zeta(3)]; %identifying Rayleigh damping coefficients
a0=a(1); %Rayleigh damping coefficients (for Mass matrix) a1=a(2); %Rayleigh damping coefficients (for Stiffness matrix) zeta(2)=a0/(2*wn(2))+a1/2*wn(2) %damping ratio for the second mode based on the obtained coefficients
for i=1:3; wd(i)=wn(i)*(1-zeta(i)^2)^0.5; %damped frequency m(i)=phi(:,i)'*M*phi(:,i); %modal mass y0(i)=phi(:,i)'*M*u0/m(i); ydot0(i)=phi(:,i)'*M*udot0/m(i);
end; k=0; for t=0:0.0001:100;
k=k+1; s=0;
for i=1:3; s=phi(:,i)*exp(-zeta(i)*wn(i)*t)*(y0(i)*cos(wd(i)*t)+ (1/wd(i))*(ydot0(i)+zeta(i)*wn(i)*y0(i))*sin(wd(i)*t))+s;
end; u1(k)=s(1); u2(k)=s(2); u3(k)=s(3); end;
80CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
t=0:0.0001:100; figure subplot(311) plot(t,u1,'linewidth',2) xlabel('time sec') ylabel('u_1') set (gca,'fontsize',18) subplot(312) plot(t,u2,'linewidth',2) xlabel('time sec') ylabel('u_2') set (gca,'fontsize',18) subplot(313) plot(t,u3,'linewidth',2) xlabel('time sec') ylabel('u_3') set (gca,'fontsize',18) %Check time response against the given initial conditions u1(1) u2(1) u3(1)
81CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
82CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Forced Vibration - Undamped MDOF System
Orthogonality
83CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Example
The 2-DOF system shown below is subjected to a single harmonic force, P1(t) = P1 cos(Ωt). Determine the steady-state response of each of the masses as a function of frequency Ω. Use the modal expansion method in solving this problem.
For a particular case of k=1,000N/m and m=10kg plot the frequency response functions.
84CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution
¾ Write the equation of motion
¾ Obtain eigenvalues and eigenvectors
85CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution
¾ Use modal expansion and transform the equation of motion to principal coordinates (uncouple the equations)
86CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution
¾ Write the uncoupled equations of motion:
Note that the transformation to principal coordinates has uncoupled the equations of motion. This is again due to the orthogonality relationships in mode shapes. As a result, we will have two independent single dof equations which can be solved in the same manner we learned in Lecture 3.
¾ Solve the uncoupled equations (based on Lecture 3):
Since, we are interested in steady state response, particular solution will be of interest. From Lecture 3 for the particular solution of an undamped single dof system to a harmonic loading, we can obtain:
87CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution
¾ Transform back the responses to physical coordinates:
88CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution
¾ Transform back the responses to physical coordinates. :
89CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
clc; clear all; close all; k=1000; % N/m m=10; % Kg w1=(k/m)^0.5; % first natural frequency w2=(5*k/(2*m))^0.5; % second natural frequency Omega=0:0.001:20; % range of excitation frequency u1=(1/3)./(1-(Omega/w1).^2)+(4/15)./(1-(Omega/w2).^2); % steady state response ratio for the first dof u2=(1/3)./(1-(Omega/w1).^2)-(1/2)*(4/15)./(1-(Omega/w2).^2); % steady state response ratio for the second dof figure subplot(211) semilogy(Omega,abs(u1),'line id h ,2) % plot frequency response in log scale xlabel('\Omega') ylabel('k|u_1|/P_1') set(gca,'fontsize',14) grid minor subplot(212) semilogy(Omega,abs(u2),'line id h ,2) % plot frequency response in log scale xlabel('\Omega') ylabel('k|u_2|/P_1') set(gca,'fontsize',14) grid minor
90CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Solution in MATLAB
First Mode Second Mode
First Mode Second Mode
91CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
Forced Vibration - Damped MDOF System
Assuming classical damping matrix
Which can be solved in the same way discussed in Lecture 3 for SDOF systems.
92CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
State-Space Representation for a MDOF System
Solve the state space equation for eigen-vectors and eigen-values:
93CVEN9840 Structural Health Monitoring 2020Dr Mehri Makki Alamdari
State-Space Representation for a MDOF System
r=1:2N
For a system with viscous damping, there will be 2N eigenvectors and 2N eigenvalues which are complex conjugates.
