Matrix Iteration Method

By Maxwell’s reciprocal theorem a12 = a21 = 5L3 / (48 EI)

Now let only the load 2mg act at station2. The deflection curve and the B-M diagram will be as shown in Fig. E8.5(b).

8.6.Matrix Iteration Method

1. Introduction: - This is the most commonly used method among the iterative methods for determining the natural frequencies (eigen values) and the corresponding mode shapes(eigen vectors)

If we are using the flexibility coefficients to write the equations of motion, then this method will lead to the lowest natural frequency of the system and the higher natural frequencies are obtained by using the orthogonal property between any two principal modes and the sweeping matrix. On the otherhand if we use the stiffness influence coefficients to write the governing equations of motion, this method will lead to the highest natural frequency and using the sweeping matrix method the remaining natural frequencies can be obtained.

2.Orthoganality between principal modes :- Let us consider the equations of motion for an

‘n’ degree-of-freedom system expressed in terms of stiffness influence coefficients. These equations in matrix form for free harmonic oscillations can be written as follows.

−ω2[mi]{xi} + [kij]{xi} = {0} ……….8.6.1.

Where [mi] = mass matrix with off-diagonal elements equal to zero, {xi} = Amplitude vector, and [ki j] = Stiffness matrix.

Or ω2 [mi]{xi} = [ki j]{xi} ………8.6.2

For one of the natural frequencies say ω = ωr, the above equation can be written as ωr2 [mi] {xi}r = [ki j] {xi}r ………..8.6.3

where {xi}r is the amplitude vector corresponding to the natural frequency ωr. Similarly for another natural frequency ωs, we can write the equation 8.6.2 as follows:

ωs2 [mi]{xi}s = [ki j] {xi}s ………8.6.4.

Pre-multiplying Eq. 8.6.3 by by transpose of {xi}s and Eq. 8.6.4. by transpose of {xi}r we get

ωr2 {xi}sT[mi] {xi}r = {xi}sT [ki j] {xi}r ...8.6.5 and ωs2 {xi}rT [mi] {xi}s = {xi}rT [ki j] {xi}s ………8.6.6 Since [mi] and [ki j] are symmetric matrices the following relations hold good.

{xi}sT[mi] {xi}r = {xi}rT[mi] {xi}s ………..8.6.7 and {xi}sT [ki j] {xi}r = {xi}rT [ki j] {xi}s ……….8.6.8 Subtracting Eq. 8.6.6 from Eq. 8.6.5 we get

(ωr2 – ωs2) {xi}rT [mi] {xi}s = 0 ………...8.6.9 If ωr ≠ ωs, then it follows that {xi}rT [mi] {xi}s = 0 ………8.6.10

And {xi}rT [ki j] {xi}s = 0 ………..8.6.11 Equations 8.6.10 and 8.6.11 define the orthogonal properties of the

normal modes of vibration.For example, for a three degree-of-freedom system Eq. 8.6.10 can be written in expanded form as follows:

{x1 x2 x3}r m1 0 0 x1 0

0 m2 0 x2 = 0 …………..8.6.12

0 0 m3 x3 s 0

After performing the matrix multiplication the above equation reduces to (x1)rm1(x1)s + (x2)rm2(x2)s + (x3)rm3(x3)s = 0 ………8.6.13

Where (x1)r, (x2)r and (x3)r are the amplitudes of vibration of masses m1, m2, and m3 respectively when the system is vibrating with natural frequency ωr and (x1)s, (x2)s, and (x3)s are the amplitudes of masses m1, m2, and m3 respectively when the system is vibrating with the natural frequency ωs.

When ωr = ωs, {xi}rT [mi] {xi}s = Mr ...8.6.14 And {xi} r [ki j] {xi}s = Kr ……….8.6.15

Where Mr and Kr are referred to as “generalized mass” and “generalized stiffness”

respectively.Mr and Kr are actually 1 x 1 matrices.

3.To determine the lowest natural frequency

:-To determine the lowest natural frequency, the governing equations for free vibrations has to be written in terms of flexibility influence coefficients as follows:

x1 = a11F1 + a12F2 + ……….a1nFn Assuming harmonic oscillations i.e. xi = xi sinωt, Eqs.8.6.17 reduces to {xi} = ω2 [aij] [mi] {xi} ………8.6.18

iteration process is started by assuming a set of displacements for {xi} and substituting on the RHS of Eq. 8.6.19. After performing the multiplication, the RHS reduces to a column vector.

This is then normalized and the procedure is repeated with the normalized column vector itself as the new estimate. The iteration process is continued till the first mode repeats itself.

The iteration process described above converges to the lowest value of ω2 so that the fundamental mode of vibration is obtained. For the next higher modes and the natural frequencies, the orthogonality principle is applied to obtain a matrix equation that does not contain the lower modes. Then the iterative process is repeated as before to get the other modes.

4. Calculation of Higher Modes :- Since the governing differential equations are linear, we can use the principle of superposition, which in this case can be stated as follows : If {xi}1, {xi}2 , …….{xi}n are the amplitude vectors corresponding to the natural frequencies ω1, ω2,

……..ωn then the linear combination of these vectors will also be a solution of the governing differential equations. That is

{xi} = C1{xi}1 + C2 {xi}2 +……….. + Cn {xi}n ………8.6.21

If we want to obtain the second mode, then we have to eliminate the first normal mode by letting C1 = 0. This is done as follows :

Pre-multiplying the above equation by {xi}1T[mi] we have

{xi}1T[mi] {xi} = C1{xi}1T[mi]{xi}1 + C2 {xi}1T [mi] {xi}2 + …….Cn{xi}1T [mi]{xi}n ………..8.6.22

Introducing the orthogonality principle (Eq. 8.6.10) in Eq. 8.6.22 we get {xi}1T [mi] {xi} = C1 {xi}1T [mi] {xi}1

By letting C1 = 0 we have {xi}1T [mi] {xi} = {0} ………8.6.23(a)

The expanded form of the above equation for a three degree-of-freedom system will be as follows:

m1 0 0 x1 0

{ x1 x2 x3 }1 0 m2 0 x2 = 0 …..8.6.23(b)

0 0 m3 x3 0

Eq. 8.6.23(b) simplifies to

(x1)1m1(x1) + (x2)1m2(x2) + (x3)1m3(x3) = 0 …………..8.6.23(c) Solving for (x1) from the above equation we get

(x2)1 m2 (x3)1 m3 Equations 8.6.24 can be written in matrix form as :

{xi} = [si j] {xi} ……….8.6.25

Since Eq.8.6.25 is the result of putting C1 = 0, the first mode of vibration is eliminated or swept out by the sweeping matrix [si j]. Therefore replacing {xi} on the right hand side of Eq.8.6.19 we get

{xi} = ω2 [bi j] [si j] {xi} ………..8.6.26

The above equation is used for the iteration process to get the second lowest mode of vibration. For obtaining the third lowest mode the following equations, which will eliminate the first and second modes are used.

(x1)1m1(x1) + (x2)1m2 (x2) + (x3)1m3(x3) = 0 ………8.6.27(a) (x1) 2m1(x1) + (x2) 2m2 (x2) + (x3) 2m3(x3) = 0………8.6.27(b)

Eq.8.6.27(b) is obtained by introducing orthogonality relationship by pre-multiplying Eq.8.6.21 by {xi}2T [mi] and setting C2 = 0.The sweeping matrix [si j] is formed by using the Eqs. 8.6.27(a) and 8.6.27(b) along with the identity

(x3) = (x3) …………..8.6.27(c)

Equations 8.6.27 (a) to (c) can be rewritten as follows:

(x1) = 0 − [(x2)1m2 / (x1)1m1](x2) − [(x3)1m3 / (x1)1m1] (x3) (x2) = −[(x1)2m1/(x2)2m2](x1)− [(x3)2m3 / (x2)2m2] (x3) and (x3) = (x3) ……….8.6.28

Equations 8.6.28 can be written in matrix form as:

{xi} = [sij] {xi} ………..8.6.29 Where [sij] is given by

0 − [(x2)1m2 / (x1)1m1] − [(x3)1m3 / (x1)1m1]

[sij] = −[(x1)2m1/(x2)2m2] 0 − [(x3)2m3 / (x2)2m2]

0 0 1

Now {xi} is replaced by [sij]{xi} on the right hand side of Eq. 8.6.19 and

The iteration process is repeated as done earlier to obtain the second lowest mode shape.

5. To Find the Highest Natural Frequency:- ..

[mj] {xi} + [kij] {xi} = {0} ………. 8.6.21 Pre-multiplying by [mi]−1 we get

..

[I] {xi} + [mi] [kij] {xi} = {0}

..

Or [I] {xi} + [dij] {xi} = {0} ………8.6.22 Where [dij] = [mi]−1 [kij]

Assuming harmonic oscillations, i.e., {xi} = {xi} sin ωt, Eq. 8.6.22 reduces to

− ω2 {xi} + [dij] {xi} = {0}

Or {xi} = ( 1 / ω2) [ d I j ] {xi} ………8.6.23 The expanded form of the above equation would be

x1 d11 d12 … d1n x1 x2 d21 d22 … d2n x2 : : : : : x n dn1 dn2 … dnn xn

Using Eq. 8.6.24, the iteration process as described above is started and this converges to the lowest value of (1 / ω2) so that the highest mode of vibration is obtained.

For the next lower modes and the natural frequencies the orthogonality principle is applied to obtain a modified matrix equation that does not contain the higher modes. The process is repeated as before.

The method is illustrated by the following examples.

Illustrative Examples on Matrix Iteration Method