1. Noise and Vibrations
(BDC4013)
Chapter 5 – Computational/Numerical Methods
Determination of Natural Frequencies and Mode Shapes
DR MUHD HAFEEZ ZAINULABIDIN
Universiti Tun Hussein Onn Malaysia

2. Necessity to Use Computational Method

In two degrees of freedom system,
solving the natural frequencies can be
conducted by simply calculating the
root of the second order polynomial.
Aω 4 + Bω 2 + C = 0
By assuming
( )
A ωn
2 2
+ Bω n + C = 0
2
Then the natural frequencies can be found
2

3. Classical Methods




Standard Matrix Iteration Method
Dunkerly’s Method
Rayleigh’s Method
Holzer’s Method
3

4. Standard Matrix Iteration
Considering a general
equation of motion
x
[ M ] { &&} + [ K ] { x} = 0
Assuming harmonic
motion
xi (t ) = X i sin(ωt )
Equation to solve
−ω 2 [ M ] { X } + [ K ] { X } = 0
4

5. Standard Matrix Iteration
(two possible solution)
−ω 2 [ M ] { X } + [ K ] { X } = 0
multiply
−ω 2 [ K ]
−1
−1
[ M ]{ X} +[ K]
−ω 2 [ K ]
[ K]
[ K]
−1
−1
multiply
−1
[ K]{ X} = 0
[ M ]{ X} +[ I]{ X} = 0
1
[ M ]{ X} = 2 { X}
ω
Converge to lowest nat freq
−ω 2 [ M ]
−1
[M]
[ M ]{ X} +[ M ]
−ω 2 [ I ] { X } + [ M ]
−1
−1
−1
[ K]{ X} = 0
[ K]{ X} = 0
[ M ] [ K ] { X } = ω2 { X }
−1
Converge to highest nat freq
5

6. Standard Matrix Iteration
(Solution procedures to obtain the lowest nat freq)
(1) Identify matrix [K] and [M]
(2) Calculate [K]-1
(3) Define the initial trial vector {X} and convergence criteria
(4) Multiply [K]-1 [M] {X} = 1/ω2{Xnew}
(5) Normalized the result {Xnew}/{X1new)
(6) Check the convergence , use for a new trial {X}
(7) When it is converged
1
= X normalized
2
ωn
6

7. Standard Matrix Iteration
(Solution procedures to obtain the highest nat freq)
(1) Identify matrix [K] and [M]
(2) Calculate [M]-1
(3) Define the initial trial vector {X} and convergence criteria
(4) Multiply [M]-1 [K] {X} = ω2{Xnew}
(5) Normalized the result {Xnew}/{X1new)
(6) Check the convergence , use for a new trial {X}
(7) When it is converged
ωn = X normalized
2
7

8. Example Problem 9-1(a)
Calculate the fundamental (lowest)
natural frequency and the
corresponding mode shapes.
k1
m1
x1
k2
m2
k1=10N/m
m1 = 1.2 kg
k2=20N/m
k3=15N/m
m2 = 2.7 kg
x2
k3
8

9. Example Problem 9-1(b)
Calculate the highest natural
frequency and the corresponding
mode shapes.
k1
m1
x1
k2
m2
k1=10N/m
m1 = 1.2 kg
k2=20N/m
k3=15N/m
m2 = 2.7 kg
x2
k3
9

10. Dunkerly’ Formula
Dunkerly’ formula is searching for the fundamental (lowest)
natural frequency.
It is based on [K]-1 multiplication
−ω 2 [ K ]
−1
[ M ]{ X} +[ I]{ X} = 0
[ K] [ M ]{ X} −
−1
1
I X } = { 0}
2 [ ]{
ω
1
−1

[ K ] [ M ] − 2 [ I ]  { X } = { 0}


ω


[ a] [ M ] −
[ K]
−1
= [ a]
1
I =0
2 [ ]
ω
10

11. Dunkerly’ Formula
(calculation procedures)
(1) Identify k11, k22, knn, m1, m2, mn
(2) Calculate natural frequency of the individual component
knn
ωnn ≈
mn
ωnn : natural frequency of a SDOF system
consisting m n and spring of stiffness k nn
(3) Predict the fundamental natural frequency of the system
1
1
1
1
≈ 2+
+L +
ωn 2 ω11 ω22 2
ωnn 2
ωn : fundamental (lowest) natural frequency
11

12. Example Problem 9-2
Predict the fundamental natural
frequency using Dunkerly method
k1
m1
x1
k1=10N/m
m1 = 1.2 kg
k2=20N/m
k3=15N/m
m2 = 2.7 kg
k2
m2
x2
k3
12

13. 13

14. 14

15. Example Problem 9-3
5 kg
7 kg
4 kg
E=207 GPa
I=12 10-6 m4
0.5m
0.5m
0.5m
0.5m
Predict the fundamental natural frequency of
the system using Dunkerly method
15

16. Solution to Problem 9.3
From the known formula for the deflection of a simply
supported beam, the flexibility influence coefficients can
be found.
Px
Deflection, v = −
(3L2 − 4 x 2 ) 0 ≤ x ≤ L
48 EI
2
3 L3
1 L3
a11 = a33 =
and a22 =
256 EI
48 EI
1  3m1 1m2 3m3  L3
≈
+
+

2
256 48 256  EI
ω1 
1  3( 5) 1( 7 ) 3( 4 ) 
23
≈
+
+

2
256 48 256  207 ×109 12 ×10 −6
ω1 
(
)(
)
ω1 = 1111.56 rad/s
16

17. Rayleigh Method


This method predicts the fundamental
(lowest) natural frequency
This method based on energy method
1
&
T = mx 2
2
T=
1
T
&
&
{ x} [ M ] { x}
2
1
V = kx 2
2
V=
1
T
x} [ K ] { x}
{
2
17

18. Rayleigh Quotient
T=
1
T
&
&
{ x} [ M ] { x}
2
V=
1
T
{ x} [ K ] { x}
2
{ x} = { X } sin(ωt )
&
{ x} = − { X } ω cos(ωt )
Tmax =
1
T
X } [ M ] { X } ω2
{
2
Vmax =
1
T
X} [ K]{ X}
{
2
Tmax = Vmax
{ X} [ K]{ X}
2
ω =
T
X} [ M ]{ X}
{
T
18

19. Rayleigh Method
(Calculation procedures)



Identify [K] and [M]
Select any trial vector mode {X}
Predict the fundamental natural
frequency based on the Rayleigh
Quotient
{ X} [ K]{ X}
T
{ X} [ M ]{ X}
T
ω2 =
19

20. Example problem 9-4
Predict the fundamental natural
frequency using Rayleigh method
k1
m1
x1
k1=10N/m
m1 = 1.2 kg
k2=20N/m
k3=15N/m
m2 = 2.7 kg
k2
m2
x2
k3
20

21. 21

22. Holzer Method
&&
I1θ1 = −kt1 (θ1 − θ 2 )
&&
I 2θ 2 = −kt1 (θ 2 − θ1 ) − kt 2 (θ 2 − θ 3 )
&&
I θ = −k (θ − θ )
3 3
t3
3
ω 2 I1Θ1 = kt1 (Θ1 − Θ 2 )
ω 2 I 2 Θ 2 = kt1 (Θ 2 − Θ1 ) + kt 2 (Θ 2 − Θ3 )
ω 2 I 3 Θ 3 = kt 3 ( Θ 3 − Θ 2 )
2
Assume
θ i = Θi cos(ωt + φ )
+
n
ω 2 I i Θi = 0
∑
i =1
22

23. Holzer Method
(calculation)
ω I1Θ1 = kt1 (Θ1 − Θ 2 )
2
ω 2 I 2 Θ 2 = kt1 (Θ 2 − Θ1 ) + kt 2 (Θ 2 − Θ3 )
ω 2 I1Θ1
Θ2 = Θ1 −
kt 1
kt 2 Θ3 = kt 2Θ 2 + kt1 (Θ 2 − Θ1 ) − ω 2 I 2 Θ 2
kt 1
ω 2 I 2Θ2
Θ3 = Θ 2 +
(Θ 2 − Θ1 ) −
kt 2
kt 2
ω 2 I1Θ1 ω 2 I 2 Θ 2
Θ3 = Θ 2 −
−
kt 2
kt 2
ω2
Θ3 = Θ 2 −
( I1Θ1 + I 2Θ2 )
kt 2
23

24. Holzer Method
(calculation procedures)


Set initial ω=0 and set the sweep increment of ω with a value Δω
Station 1:
X1=1 (or Θ1=1), calculate M1=ω2m1X1 (or ω2I1Θ1)

Station 2:
Calculate X2 (or Θ 2), calculate M2=M1+ ω2m2X2 (or ω2I2Θ2)

Station 3:
Calculate X3 (or Θ 3), calculate M3=M2+ ω2m3X3 (or ω2I3Θ3)

Station n:
Calculate Xn (or Θ n), calculate Mn=Mn-1+ ω2mnXn (or ω2InΘn)
24

25. Example problem 9-5
I1=2 kg m2
I2=4 kg m2
Calculate the natural
frequencies and mode
shapes
Kt=4 MNm/rad
25

26. 26

27. 27

28. Holzer Method
(summary calculation)
Torsion
Translation
Θ1 = 1
X1 = 1
ω 2 I1Θ1
Θ2 = Θ1 −
kt 1
ω 2 m1 X 1
X 2 = X1 −
k1
ω2
Θ3 = Θ 2 −
( I1Θ1 + I 2Θ 2 )
kt 2
ω2
X3 = X2 −
( m1 X 1 + m2 X 2 )
k2
ω 2  i −1

Θi = Θi −1 −
I k Θk ÷
∑ 
kti −1  k =1

i = 2,3,L n
ω 2  i −1

X i = X i −1 −
mk X k ÷
∑
ki −1  k =1


i = 2,3,L n
28

29. Example problem 9-6
I1=2 kg m2
I2=4 kg m2
Calculate the natural
frequencies and mode
shapes
I3=2 kg m2
kt1=3 MNm/rad
Kt2=2 MNm/rad
29