Modal and Harmonic Analysis With ANSYS

David Herrin, Ph.D.University of Kentucky

Modal and Harmonic Analysis using ANSYS

Modal/Harmonic Analysis Using ANSYS

ME 599/699 Vibro-Acoustic Design

Dept. of Mechanical Engineering University of Kentucky

Modal Analysis

g Used to determine the natural frequencies and mode shapes of a continuous structure




Review of Multi DOF Systems

g Expressed in matrix form as

M1 M2

K2

C2

K1

C1

K3

C3

x1 x2

F1 F2

[ ]{ } [ ]{ } [ ]{ } { }FuKuCuM =++ &&&




Review of Multi DOF Systems

[ ]{ } [ ]{ } [ ]{ } { }FuKuCuM =++ &&&

g The mass, damping and stiffness matrices are constant with time

g The unknown nodal displacements vary with time




Modal Analysis

g A continuous structure has an infinite number of degrees of freedom

g The finite element method approximates the real structure with a finite number of DOFs

g N mode shapes can be found for a FEM having N DOFs

g Modal Analysis

ü Process for determining the N natural frequencies and mode shapes




Modal Analysis

g Given “suitable” initial conditions, the structure will vibrate

ü at one of its natural frequencies

ü the shape of the vibration will be a scalar multiple of a mode shape

g Given “arbitrary” initial conditions, the resulting vibration will be a

ü Superposition of mode shapes




Modal Analysis

g Determines the vibration characteristics (natural frequencies and mode shapes) of a structural components

g Natural frequencies and mode shapes are a starting point for a transient or harmonic analysis

ü If using the mode superposition method




Mode Shape of a Thin Plate (240 Hz)




Mode Extraction Methods

g Subspace

g Block Lanczos

g PowerDynamics

g Reduced

g Unsymmetric

g Damped and QR damped (Include damping)




Steps in Modal Analysis

g Build the model

ü Same as for static analysis

ü Use top-down or bottom-up techniques

g Apply loads and obtain solution

ü Only valid loads are zero-value displacement constraints

ü Other loads can be specified but are ignored

g Expand the modes and review results




360 inches

IZZ = 450 in4

A = 3 in2

H = 5 in

In-Class Exercise

E = 30E6 psi

2

3

lb s8.031 4

in inE

ρ = −




In-Class Exercise

g Set element type to BEAM3

g Set the appropriate real constants and material constants

g Create Keypoints at the start and end of the beam and a Line between them




Mesh the Line and Apply B.C.s

g Set the meshing size control of the line to 5

g Fix the Keypoint at the right end of the beam




Set Solution Options

g Change the analysis type to Modal

ü Solution > Analysis Type > New Analysis

ü <Modal>

g Set the analysis options

ü Solution > Analysis Options

ü Extract 10 mode <OK>

ü Enter <1500> for the ending frequency





g At this point, you have told ANSYS to find a particular quantity of modes and to look within a particular frequency range. If ANSYS finds that quantity before it finishes the frequency range, it will stop the search. If ANSYS does not find that quantity before finishing the frequency range, then it will stop the search.





g Solve the load set

g ANSYS generates a substep result for each natural frequency and mode shape




Postprocessing

g List results summary

ü General Postproc > List Results > Results Summary

g Read results for a substep

ü General Postproc > Read Results > First Set

ü Plot deformed geometry

ü General Postproc > Read Results > Next Set



Theoretical Modal Analysis




Structural Vibration at a Single DOF

g At a single degree of freedom

( )tu ωφ= cos

DOF nodal a ofnt displaceme=u

amplitude=φ




Structural Vibration for the Entire Structure

{ } { } ( )tu ωφ= cos

{ } ntsdisplaceme nodal ofVector =u

{ } DOFeach for amplitudes ofvector =φ




Mode Shapes

g The vector of amplitudes is a mode shape

2φ 3φ 4φ 5φ6φ

7φ8φ

g Use subscript i to differentiate mode shapes and natural frequencies

{ } { } ( )tu ii ωφ= cos




Appropriate Initial Conditions

g If the I.C.s are a scalar (a) multiple of a specific mode shape then the structure will vibrate in the corresponding mode and natural frequency

{ }ia φ=ICs




Determining Natural Frequencies

g Consider a multi DOF system

[ ]{ } [ ]{ } { }0=+ uKuM &&

g Note that the system is

ü Undamped

ü Not excited by any external forces




Modal Analysis

{ } { } ( )tu ii ωφ= cos

g If the system vibrates according to a particular mode shape and frequency

{ } ii shape mode=φ

ii frequency natural=ω




Modal Analysis

{ } { } ( )tu iii ωφω sin−=&

g First derivative (Velocity)

g Second derivative (Acceleration)

{ } { } ( )2 cosi iiu t= −ω φ ω&&




Modal Analysis

[ ] [ ]( ){ } { }2 0i iM K− ω + φ =

g Plug velocity and acceleration into the equation of motion

g Two possible solutions

{ } { }0i

φ =

[ ] [ ]( )2det 0iM K− ω + =




The Eigen Problem

g The Eigen Problem

[ ] [ ]( )2det 0iM K− ω + =

λg Eigenvalue

{ }xg Eigenvector

[ ]{ } [ ]{ }xIxA λ=




Modal Analysis – An Eigen Problem

[ ] [ ]( ){ } [ ]2 0iM K− ω + φ =

g Natural frequencies are eigenvalues

g The mode shapes are eigenvectors

[ ]{ } [ ]{ }2iK Mφ = ω φ

2iω

{ }φ

[ ] [ ]{ } [ ]{ }φω=φ− IKM i21




Modal Analysis – An Eigen Problem

g Solving the eigen problem

[ ] [ ]( ){ } 0A I x− λ =

[ ] [ ]( )det 0A I− λ =

[ ]{ } [ ]{ }xIxA λ=

[ ] [ ]{ } [ ]{ }φω=φ− IKM i21




Example Eigen Problem

1 1

2 2

2 33 2 i

x xx x

= λ

2 3 1 0det 0

3 2 0 1i

− λ =




Find Eigenvalues

2 3det 0

3 2 − λ

= − λ ( )( ) 22 2 9 4 5− λ − λ − = λ − λ −

g Eigenvalues

1 1λ = −

2 5λ =




For Eigenvector 1

1 1λ = −

( ) 1

2 1

2 3 1 0 01

3 2 0 1 0xx

− − =

1

2 1

2 1 3 03 2 1 0

xx

+ = +




For Eigenvector 1

1

2 1

3 3 03 3 0

xx

=

2 1x x= −

g x1 can be any value

1

2 1

1 .70711 .7071

xx

= = − −




For Eigenvector 2

2 5λ =

( ) 1

2 2

2 3 1 0 05

3 2 0 1 0xx

− =

1

2 2

2 5 3 03 2 5 0

xx

− = −




For Eigenvector 2

1

2 2

3 3 03 3 0

xx

− = −

2 1x x=

g x2 can be any value

1

2 2

1 .70711 .7071

xx

= =




Example

g Equations of motions

M1 M2

K1 K2

x1 x2

1 1 1 1 1 2 0M x K x K x+ − =&&

2 2 2 2 1 2 1 1 0M x K x K x K x+ + − =&&

1 5 kgM =

2 10 kgM =

1 800 N/mK =

2 400 N/mK =




Matrix Form

1 1 1 1 1

2 2 1 1 2 2

0 00 0

M x K K xM x K K K x

− + = − +

&&&&

1 1

2 2

5 0 800 800 00 10 800 1200 0

x xx x

− + = −

&&&&




Set Up Eigen Problem

[ ] [ ]{ } { }1 2iM K− φ = ω φ

[ ] [ ]1 160 16080 120

M K− −

= −

160 160 1 0det 0

80 120 0 1 −

− λ = −




Solve Eigen Problem

160 160det 0

80 120 − λ −

= − − λ 1 1 5.011 rad/sω = λ =

2 2 15.965 rad/sω = λ =

1

2 1

0.76450.6446

φ = φ

1

2 1

0.86010.5101

φ − = φ

1

5.011.7975 Hz

2f = =

π

2

15.9652.5409 Hz

2f = =

π




Weighted Orthogonality

[ ] 022

1

12

1 =

φφ

φφ

MT

[ ] 022

1

12

1 =

φφ

φφ

KT




Modal Vector Scaling

[ ] 112

1

12

1 =

φφ

φφ

MT

g Unity Modal Mass




Modal Vector Scaling

164467645

10005

64467645

11

=

XX

XX

T

.

...

g Unity Modal Mass

376.=X

=

φφ

242287

12

1

.

.




Mode 1

M1 M2

K1 K2

0.7645 0.6446




Mode 2

M1 M2

K1 K2

-0.8601 0.5101


Harmonic Response using ANSYS




Harmonic Response Analysis

g Solves the time-dependent equations of motion for linear structures undergoing steady-state vibration

g All loads and displacements vary sinusoidally at the same frequency

1sin( )iF F t= ω + φ

2sin( )jF F t= ω + φ




Harmonic Response Analysis

g Analyses can generate plots of displacement amplitudes at given points in the structure as a function of forcing frequency




360 inches

IZZ = 450 in4

A = 3 in2

H = 5 in

In-Class Exercise

E = 30E6 psi

2

3

lb s8.031 4

in inE

ρ = −




In-Class Exercise

g Set element type to BEAM3

g Set the appropriate real constants and material constants

g Create Keypoints at the start and end of the beam and a Line between them




Mesh the Line and Apply B.C.s

g Set the meshing size control of the line to 5

g Fix the Keypoint at the right end of the beam





g Change the analysis type to Modal

ü Solution > Analysis Type > New Analysis

ü <Modal>

g Set the analysis options

ü Solution > Analysis Options

ü Extract 10 mode <OK>

ü Enter <1500> for the ending frequency





g At this point, you have told ANSYS to find a particular quantity of modes and to look within a particular frequency range. If ANSYS finds that quantity before it finishes the frequency range, it will stop the search. If ANSYS does not find that quantity before finishing the frequency range, then it will stop the search.





g Solve the load set

g ANSYS generates a substep result for each natural frequency and mode shape




Postprocessing

g List results summary

ü General Postproc > List Results > Results Summary

g Read results for a substep

ü General Postproc > Read Results > First Set


ü General Postproc > Read Results > Next Set





Forced Response

g Apply a 1.0 N load at the left end of the beam

g Set the Analysis Options

ü Set the solution method to “Full”

ü Set the DOF printout format to “Amplitude and phase” <OK>

ü Set the tolerance to 1e-8 <OK>

g New Analysis > Harmonic




Forced Response

g Set the frequency substeps

ü Solution > Load Step Opts – Time Frequency > Freq and Substeps

ü Set the Harmonic Frequency Range to between 0 and 50 Hz

ü Set the number of substeps to 100

ü Set to Stepped




Forced Response

g Set the damping

ü Solution > Load Step Opts – Time Frequency > Damping

ü Set the Constant Damping Ratio to 0.01

g Solve the model




Forced Response

g Enter time history postprocessor

g Define a variable for UY at the leftmost node

g List and graph the variable

g Add a vertical truss member at the center of the beam having an area of 10.

Modal and Harmonic Analysis With ANSYS

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