Download presentation
Presentation is loading. Please wait.
Published byChad Dorsey Modified over 7 years ago
1
Chapter 7 Determination of Natural Frequencies and Mode shapes
2
Chapter Outline 7.1 Introduction 7.2 Dunkerley’s Formula
7.3 Rayleigh’s Method 7.4 Holzer’s Method 7.5 Matrix Iteration Method 7.6 Jacobi’s Method 7.7 Standard Eigenvalue Problem
3
7.1 Introduction 7.1
4
7.1 Introduction Computing the natural frequencies and modes by solving a nth degree polynomial equation can be tedious In this chapter we shall consider several other methods: Dunkerley’s formula Rayleigh’s method Holzer’s method Matrix iteration method Jacobi’s method
5
7.2 Dunkerley’s Formula 7.2
6
7.2 Dunkerley’s Formula It gives the approx. value of the fundamental frequency of a composite system. Consider the following general n DOF system: For a lumped mass system with diagonal mass matrix, the equation becomes:
7
7.2 Dunkerley’s Formula i.e. Expanding:
8
7.2 Dunkerley’s Formula Let the roots of this equation be 1/ω12, 1/ω22,…, 1/ωn2. Thus Equating coefficients of (1/ω2)n-1 in (E.1) and (E.2): In most cases,
9
7.2 Dunkerley’s Formula Thus Can also be written as
where ωin=(1/aiimi)1/2=(kii/mi)1/2
10
7.2 Dunkerley’s Formula Example 7.1
Fundamental Frequency of a Beam Estimate the fundamental natural frequency of a simply supported beam carrying 3 identical equally spaced masses, as shown below.
11
7.2 Dunkerley’s Formula Example 7.1 Fundamental Frequency of a Beam
Solution We have Since m1=m2=m3=m ,
12
7.3 Rayleigh’s Method 7.3
13
7.3 Rayleigh’s Method Based on Rayleigh’s Principle
Kinetic and potential energies of an n-DOF discrete system: Assume harmonic motion to be where is the mode shape and ω is the natural frequency
14
7.3 Rayleigh’s Method Maximum KE: Maximum PE:
For a conservative system, Tmax=Vmax
15
7.3 Rayleigh’s Method Properties of Rayleigh’s Quotient
has a stationary value when is in the vicinity of any eigenvector Proof:
16
7.3 Rayleigh’s Method Properties of Rayleigh’s Quotient
If normal modes are normalized,
17
7.3 Rayleigh’s Method Properties of Rayleigh’s Quotient
where 0(ε) is an expression in ε of the 2nd order or higher. i.e. differs from by a small quantity of the 2nd order. i.e. Rayleigh’s quotient has a stationary value in the neighborhood of an eigenvector.
18
7.3 Rayleigh’s Method Properties of Rayleigh’s Quotient
The stationary value is a minimum value in the neighborhood of To see this, let r = 1.
19
7.3 Rayleigh’s Method Properties of Rayleigh’s Quotient
In general, Rayleigh’s quotient is never lower than the 1st eigenvalue. Similarly we can show that Rayleigh’s quotient is never higher than the highest eigenvalue.
20
7.3 Rayleigh’s Method Computation of Fundamental Natural Frequency
Rayleigh’s quotient can be used to approximate ω1. Select a trial vector and substitute into This will yield a good estimate of The closer resembles the true mode , the more accurate is the estimated ω1.
21
7.3 Rayleigh’s Method Example 7.2 Fundamental Frequency of a Three-Degree-of-Freedom System Estimate the fundamental frequency of vibration of the system as shown. Assume that m1=m2=m3=m, k1=k2=k3=k, and the mode shape is
22
7.3 Rayleigh’s Method Example 7.2 Fundamental Frequency of a Three-Degree-of-Freedom System Solution Stiffness matrix Mass matrix Substitute the assumed mode shape into
23
7.3 Rayleigh’s Method Fundamental Frequency of Beams and Shafts
Static deflection curve is used to approximate the dynamic deflective curve. Consider a shaft carrying several masses as shown below.
24
7.3 Rayleigh’s Method Fundamental Frequency of Beams and Shafts
Potential energy of the system is strain energy of the deflected shaft, which is the work done by the static loads. For free vibration, max kinetic energy due to the masses is Equating Vmax and Tmax,
25
7.4 Holzer’s Method 7.4
26
7.4 Holzer’s Method A trial-and-error scheme to find natural frequencies of systems A trial frequency is first assumed, and a solution is found when the constraints are satisfied. Requires several trials The method also gives mode shapes
27
7.4 Holzer’s Method Torsional Systems
Consider the undamped torsional semidefinite system shown below. Equations of motion
28
7.4 Holzer’s Method Torsional Systems
Since the motion is harmonic, θi=Θicos(ωt+φ) Summing these equations gives This states that the sum of the inertia torques of the system must be zero. The trial freq must satisfy this requirement.
29
7.4 Holzer’s Method Torsional Systems is arbitrarily chosen as 1.
Substitute these values into to see whether the constraints are satisfied. If not, repeat the process with a new trial value of ω. These equations can be generalized for a n-disc system as follows:
30
7.4 Holzer’s Method Torsional Systems
The graph below plots the torque Mt applied at the last disc against the chosen ω. The natural frequencies are the ω at which Mt=0. The amplitudes (i=1,2,…,n) are the mode shapes of the system
31
7.4 Holzer’s Method Example 7.4 Natural Frequencies of a Torsional System Solution The arrangement of the compressor, turbine and generator in a thermal power plant is shown below. Find the natural frequencies and mode shapes of the system.
32
7.4 Holzer’s Method Example 5.4 Natural Frequencies of a Torsional System Solution This is an unrestrained torsional system. The table below shows its parameters and the sequence of computations.
33
7.4 Holzer’s Method Example 5.4 Natural Frequencies of a Torsional System Solution Mt3 is the torque to the right of the generator, which must be zero at the natural frequencies. Closely-spaced trial values of ω are used in the vicinity of Mt3=0 to obtain accurate values of the 1st two flexible mode shapes, as shown.
34
7.4 Holzer’s Method Spring-Mass Systems
Holzer’s method is also applicable to vibration analysis of spring-mass systems. Equations of motion: For harmonic motion, xi(t)=Xicosωt where Xi is the amplitude of mass mi. Thus
35
7.4 Holzer’s Method Spring-Mass Systems
The resultant force applied to the last (nth) mass can be computed as follows: Repeat for several other trial frequencies ω. Plot a graph of F vs ω. The natural frequencies are those ω that give F=0.
36
7.5 Matrix Iteration Method
37
7.5 Matrix Iteration Method
The method assume that the natural frequencies are distinct and well separated. Procedure Select a trial vector Premultiply it by the dynamical matrix [D]. Normalize the resultant column vector. Repeat step 2 and 3 until the successive normalized vectors converge.
38
7.5 Matrix Iteration Method
Proof: Expansion theorem is a known vector selected arbitrarily. are constant vectors because they depend on the system properties. ci are unknown numbers to be determined. Premultiplying by [D]:
39
7.5 Matrix Iteration Method
Proof: Recall: Hence Repeating the process for r iterations: Since ω1<ω2<…<ωn, if r is large we have:
40
7.5 Matrix Iteration Method
Proof: The only significant we have on the RHS is: Since ω1 can be found by
41
7.5 Matrix Iteration Method
Discussion: A finite number of iterations is sufficient to obtain a good estimate of ω1. Actual no. of iterations depend on how close resembles Advantage: Computational errors will not yield incorrect results. The method fails if is exactly proportional to one of the modes
42
7.5 Matrix Iteration Method
Convergence to the Highest Natural Frequency To obtain ωn and the corresponding Select an arbitrary and premultiply by [D]-1 to obtain an improved trial vector The sequence of trial vectors will converge to the highest normal mode Constant of proportionality in this case is ω2 instead of 1/ ω2
43
7.5 Matrix Iteration Method
Computation of Intermediate Natural Frequencies Once ω1 and is found, we can find the higher natural frequencies. Because any premultiplied by [D] would lead to the largest eigenvalue, it is necessary to remove the largest eigenvalue from [D]. Succeeding λi and can be obtained by eliminating the root λ1 from the characteristic equation |[D] – λ[I]|=0
44
7.5 Matrix Iteration Method
Computation of Intermediate Natural Frequencies Procedure: To find normalize wrt mass matrix: Deflated matrix [Di] is constructed as: Next the iterative scheme is used, where is an arbitrary trial eigenvector.
45
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Find the natural frequencies and mode shapes of the system as shown for k1=k2=k3=k and m1=m2=m3=m by the matrix iteration method.
46
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution Flexibility matrix [a]=[k]-1= Dynamical matrix is Eigenvalue problem:
47
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution 1st natural frequency: Assume , hence By making the first element equal to unity we obtain
48
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution 1st natural frequency: Subsequent trial eigenvector can be obtained from Corresponding eigenvalues are given by where is the 1st component of before normalization.
49
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution The various λi and are shown: The mode shape and natural frequency converged in 8 iterations.
50
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution 2nd natural frequency: Deflated matrix Let the normalized vector where α must be such that
51
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution 2nd natural frequency: α= m-1/2 , hence
52
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution 2nd natural frequency: Let By using the iterative scheme, we obtain
53
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution 2nd natural frequency: Continuing the procedure, Hence λ2= , ω2=
54
7.5 Matrix Iteration Method
Example 7.5 Natural Frequencies of a Three-Degree-of-Freedom System Solution 3rd natural frequency: Use a similar procedure as before. Before computing [D3], need to normalize
55
7.6 Jacobi’s Method 7.6
56
7.6 Jacobi’s Method Produces all the eigenvalues and eigenvectors of matrix [D] simultaneously. [D]=[dij] is a real symmetric matrix of order n x n. [D] has only real eigenvalues. There exists a real orthogonal matrix [R] such that [R]T[D][R] is diagonal. The diagonal elements are the eigenvalues, and the columns of [R] are the eigenvectors.
57
7.6 Jacobi’s Method [R] is generated as a product of several rotation matrices of the form where all elements other than those in column and row i and j are identical with those of the identity matrix [I].
58
7.6 Jacobi’s Method If the sine and cosine entries appear in positions (i,i), (i,j), (j,i) and (j,j), the corresponding elements of [R]T[D][R] is as follows: If θ is chosen to be , then Successive matrices converge to the required diagonal form.
59
7.6 Jacobi’s Method Example 7.6
Eigenvalue Solution Using Jacobi Method Find the eigenvalues and eigenvectors of the matrix using Jacobi’s method.
60
7.6 Jacobi’s Method Example 7.6
Eigenvalue Solution Using Jacobi Method Solution First try to reduce d23=2 to zero.
61
7.6 Jacobi’s Method Example 7.6
Eigenvalue Solution Using Jacobi Method Solution Next try to reduce d13’= to zero.
62
7.6 Jacobi’s Method Example 7.6
Eigenvalue Solution Using Jacobi Method Solution Next try to reduce d12’’= to zero.
63
7.6 Jacobi’s Method Example 7.6
Eigenvalue Solution Using Jacobi Method Solution Assume that all the off-diagonal terms in [D”’] are close to zero. Hence the eigenvalues are , and The corresponding eigenvectors are given by
64
7.7 Standard Eigenvalue Problem
65
7.7 Standard Eigenvalue Problem
The eigenvalue problem can be written in the form of a standard eigenvalue problem Procedure: Use Choleski decomposition and express [k] as: [k]=[U]T[U] where [U] is an upper triangular matrix. Eigenvalue problem:
66
7.7 Standard Eigenvalue Problem
Procedure: Premultiplying by ([U]T)-1, Define a new vector Eq.A becomes Its solution yields λi and Apply inverse transformation to find the desired eigenvectors.
67
7.7 Standard Eigenvalue Problem
Choleski Decomposition Any symmetric and positive definite matrix [A] of order n x n can be decomposed uniquely. [A]=[U]T[U] where
68
7.7 Standard Eigenvalue Problem
Choleski Decomposition If the inverse of [U] is denoted as [αij], the elements αij can be determined from [U][U]-1=[I] which gives Thus the inverse of [U] is also an upper triangular matrix.
69
7.7 Standard Eigenvalue Problem
Example 7.7 Decomposition of a Symmetric Matrix Decompose the matrix into the form [A]=[U]T[U]
70
7.7 Standard Eigenvalue Problem
Example 7.7 Decomposition of a Symmetric Matrix Solution
71
7.7 Standard Eigenvalue Problem
Example 7.7 Decomposition of a Symmetric Matrix Solution Since uij=0 for i>j, we have
Similar presentations
© 2025 SlidePlayer.com Inc.
All rights reserved.