An Introduction to computer modelling

Modal Analysis

Dynamics

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Introduction to computer modelling

Contents

1) Objectives

2) Requirements

3) Procedure

4) Equations Of Motion

a) Formulate the mass and stiffness matrices

b) Select subset of the equation representing a 3 D.O.F system

c) Formulate mass and stiffness matrices for the 3 D.O.F system

5) Matrix iteration for 3 degree of freedom system

6) Matlab simulation

a) 3 Degree of Freedom system

b) 5 degree of freedom system

7) Discussions

8) Conclusion

9) Bibliography

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Introduction to computer modelling

1. Objectives

Set up a mathematical model of a sport case

Use this model to predict the dynamic behaviour of the car using two different

methods

a) Matrix Iteration

b) Computer Package

2. Requirements

Access to the computer package Matlab

3. Procedure

1. Formulate the stiffness and mass matrices of the modal

2. Select a subset of the equations of motion representing a 3 degree of freedom model

and formulate the mass and stiffness matrices for this case too.

3. Carry out matrix iteration to find mode frequencies and the mode shapes of the 3

degree of freedom system.

4. Use the computer package matlab to find eigenvalues and eigenvectors of

i. The 3 degree of freedom system and

ii. The full 5 degree of freedom system

Parameters:

Mw=35 kg Ks=40kN/m

Mb=320kg Kt=400kN/m

Me=100kg Ke=300kN/m

Ib=250kgm2 Cs=1000Ns/m

a=1.3m b=0.4m

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Introduction to computer modelling

4. Equations of motion

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Introduction to computer modelling

a. Formulate the mass and stiffness matrix

Mass matrix

Xw1 Xw2 Xe Xb θb Xw1 Mw 0 0 0 0

Xw2 0 Mw 0 0 0

Xe 0 0 Me 0 0

Xb 0 0 0 Mb 0

θb 0 0 0 0 Ib

Stiffness matrix

Xw1 Xw2 Xe Xb θb Xw1 ks+kt 0 0 -ks -0.5aks

Xw2 0 ks+kt 0 -ks 0.5aks

Xe 0 0 ke+ke -ke-ke 0

Xb -ks -ks -ke-ke ks+ks+ke+ke 0

θb -0.5aks 0.5aks 0 0 (a2ks+b

2ke)/2

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Introduction to computer modelling

b. Select subset of the equation representing a 3 D.O.F system

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Introduction to computer modelling

c. Formulate mass and stiffness matrices for the 3 D.O.F system

Mass Matrix

Xe Xb θb Xe Me 0 0

Xb 0 Mb 0

θb 0 0 Ib

Stiffness Matrix

Xe Xb θb Xe 2ke -2ke 0

Xb -2ke ks+ks+ke+ke 0

θb 0 0 (a2ks+b

2ke)/2

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Introduction to computer modelling

5. Matrix iteration for 3 degree of freedom system

Eigen value

Assuming no damping in the system and no forcing functions the equation of motion becomes:

We will perform matrix iteration for the following cases:

1. ,known as dynamics flexibility matrix for lowest eigenvalues and

eigenvectors.

2. known as dynamics flexibility matrix for highest eigenvalues and

eigenvectors.

3. A sweeping matrix”[S]” , to find the intermediate eigen value and eigenvectors.

Eigen values give the natural frerquency

Eigen Vectors give the mode shape

Matrix iteration

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Introduction to computer modelling

The lowest eigen values and eigenvector

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Introduction to computer modelling

Because

, therefore the eigenvalue ω1

2=

= 188.306 (rad/s)2

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Introduction to computer modelling

The highest eigenvalue and Eigenvector

The eigenvalue

The intermediate eigenvalue and eigenvector

To find the values the sweeping matrix was needed.

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Introduction to computer modelling

, therefore the eigenvalue ω3

2=

= 231.20 (rad/s)2

The obtained values in matrix form

1. Matrix of eigenvalues

Matrix of eigenvectors

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Introduction to computer modelling

6. Matlab Simulation

a. 3 degree of freedom system

K = 600000 -600000 0 -600000 680000 0 0 0 57800 M = 100 0 0 0 320 0 0 0 250 Minv 0.0100 0 0 0 0.0031 0 0 0.0040 A=Minv*K A = 1.0e+003 * 6.0000 -6.0000 0 -1.8750 2.1250 0 0 0 0.2312 >> [V,D]=eig(A) V = 0.9517 0.7183 0 -0.3071 0.6957 0 0 0 1.0000 D = 1.0e+003 * 7.9360 0 0 0 0.1890 0 0 0 0.2312

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Introduction to computer modelling

b. Full 5 degree of freedom system

K = 440000 0 0 -40000 -26000 0 440000 0 -40000 26000 0 0 600000 -600000 0 -40000 -40000 -600000 680000 0 -26000 26000 0 0 57800 M = 35 0 0 0 0 0 35 0 0 0 0 0 100 0 0 0 0 0 320 0 0 0 0 0 250 >> Minv=inv(M) 0.0286 0 0 0 0 0 0.0286 0 0 0 0 0 0.0100 0 0 0 0 0 0.0031 0 0 0 0 0 0.0040 >> A=Minv*K A = 1.0e+004 * 1.2571 0 0 -0.1143 -0.0743 0 1.2571 0 -0.1143 0.0743 0 0 0.6000 -0.6000 0 -0.0125 -0.0125 -0.1875 0.2125 0 -0.0104 0.0104 0 0 0.0231 >> [V,D]=eig(A) V = 0.7068 0.7071 0.0745 0.0599 -0.0640 0.7068 -0.7071 0.0745 -0.0599 -0.0640 0.0183 -0.0000 -0.9471 -0.0000 -0.7144 -0.0201 0.0000 0.3032 -0.0000 -0.6939 -0.0000 -0.0119 -0.0000 0.9964 -0.0000 D = 1.0e+004 * 1.2604 0 0 0 0 0 1.2584 0 0 0 0 0 0.7921 0 0 0 0 0 0.0219 0 0 0 0 0 0.0172

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Introduction to computer modelling

7. Discussions

Natural frequencies and their mode shapes for a 3 degree of freedom system using matrix iteration

The highest eigenvalue (natural frequency) for the 3 D.O.F,

Corresponding mode shape

The intermediate eigenvalue (natural frequency) for the 3 D.O.F

ω3=15.2(rad/s)=2.42Hz

Corresponding mode shape

The lowest eigenvalue (natural frequency) for the 3 D.O.F,

ω3=13.72rads

-1=2.18Hz

Corresponding mode shape

0.32

1

1

0.01

0.02

1

0.97

0.02

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Introduction to computer modelling

Natural frequencies and their mode shapes For a 3 degree of freedom system using matlab solution

The highest eigenvalue (natural frequency) for the 3 D.O.F,

Corresponding mode shape

The intermediate eigenvalue (natural frequency) for the 3 D.O.F

ω3=15.2(rad/s)=2.42Hz

Corresponding mode shape

The lowest eigenvalue (natural frequency) for the 3 D.O.F,

ω3=13.74rads

-1=2.18Hz

Corresponding mode shape

0.3071

0.9517

1

0.7183

0.6957

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Introduction to computer modelling

The highest natural frequency and its mode shape for full 5 degree of freedom system

using matlab solution

The highest eigenvalue (natural frequency) for the 5 D.O.F,

Corresponding mode shape

Checking accuracy by performing hand calculations

Mass Stiffness Results %error for matlab

%error for iteration ω2=k/m Matlab iteration

Mb 2ks 220 231.2 231.2 7.5% 7.5%

Me 2ke 6000 7936 7920 32.3% 32.3

0.7068 0.7068

0.0201

0.0183

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Introduction to computer modelling

Assumptions

No external force is applied i.e. F=0 and that the damping is neglected which leads a trivial

solution

Mass of spring are not taken into account which is a very small value and will not effect are

results.

The bodies are assumed to be rigid i.e. flexibility of bodies are not taken into account.

8. Conclusions

The results of matrix iteration and matlab solution are approximately the same for the Eigen

values i.e. the highest, intermediate and lowest natural frequencies, the difference in values is

no more than 0.5-1%.

The 3 D.O.F equation helped to solve it by matrix iteration as for a full system the method

would have been non applicable.

The matlab solution were very reliable and the process was very quick, and is very helpful for

high order multiple degree of freedom systems.

9. Bibliography

Modal testing, theory and

analysis

Ewins D.J Research Studies Press