finite element and numerical methods in structural dynamics · ref: dynamics of structures, theory...

Finite Element and Numerical Finite Element and Numerical

Methods in Structural Methods in Structural

Dynamics Dynamics

S M ShiyekarS M ShiyekarProfessor Civil Engineering Department

Fundamentals of Structural Dynamics and Applications to Earthquake Engineering

Sanjay Ghodawat Group of Institutes, Faculty of Engineering

7 to 12 Dec 2015

S M Shiyekar, Professor Civil Department

Degrees of Freedom

Independent displacements, rotations

that specify completely the displaced or

deformed position and orientation of the

node in a body or system.


Dynamics of Single Degree of

Freedom (SDOF) System

Static Force

P ku=

( )P t ku≠

Dynamic Force


Governing Differential Equation

2

( )

( ) N et force

( ) m ass x acceleration

P t ku

P t ku

P t ku

≠

− =

− =

2

2

d u( )

d t

(

)

)

(

P t ku

P t ku

u ku P t

mu

m

− =

−

+ =

= &

&&

&


Free Vibration

0mu ku+ =&&


Natural FrequencyEvery periodic motion is governed by

2 2

( ) cos( ) sin( )

sin( ) cos( )

cos( ) sin( )

( ) cos( ) sin( ) (Fourier)

n n

n n n n

n n n n

u t A t B t

u A t B t

u A t B t

u t A B

ω ω

ω ω ω ω

ω ω ω ω

θ θ

= +

= − +

= − −

= +

&

&&

2 2

2

2

2

2

cos( ) sin( )

( ) 0

( )

,

n n n n

n

n

n

n n

u A t B t

u u

m u ku

k m

k k

m m

ω ω ω ω

ω

ω

ω

ω ω

= − +

= −

− + =

=

= =

&&

&&


2 2

2 2

2

( ) cos( ) sin( )

( ) cos( ) sin( )

sin( ) cos( )

cos( ) sin( )

cos( ) sin( )

n n

n n n n

n n n n

n n n n

u t A B

u t A t B t

u A t B t

u A t B t

u A t B t

u u

θ θ

ω ω

ω ω ω ω

ω ω ω ω

ω ω ω ω

ω

= +

= +

= − +

= − −

= − +

= −

&

&&

&&

2

2

2

2

( ) 0

( )

,

n

n

n

n n

u u

m u ku

k m

k k

m m

ω

ω

ω

ω ω

= −

− + =

=

= =

&&


SDOF System Natural Frequency

Rayleigh-Ritz Method

{ }

{ }

0

0

( ) sin( ) : displacement

( ) cos( ) : velocity

n

n n

u t u t

u t u t

ω

ω ω

=

=&


SDOF System Natural Frequency

Rayleigh-Ritz Method

{ }

{ }

0

0

( ) sin( ) : displacement

( ) cos( ) : velocity

n

n n

u t u t

u t u t

ω

ω ω

=

=&

Strain energy stored in spring =

Kinetic energy of system =

SOE

E

2 2 2

0 0

2

Kinetic energy of system =

maximum maximum

1 1

2 2

;

KO

SO KO

n

n n

E

E E

ku mu

k k

m m

ω

ω ω

=

=

= =


Continuum System: Cantilever Beam by Rayleigh-Ritz Method

Cantilever Beam: Vibrating in SHM

[ ]

( )

0

2

2

L 2

0

2 2

( , ) sin( ) ( )

''( )

MStrin Energy in Bending =

2

[ ''( )]

n

L L

u x t z t x

d yM EI EIu x

dx

dx

EI

EI z x EI

ω ψ

ψ

=

= =

∫

( )( )

( )

2

2

2 0

2

2

20 2

0

0 0

22

0

2 2

0

0

[ ''( )][ ''( )]

2 2

1 1( ) [ ( )]

2 2

( )[ ''( )]

( )[ ( )]

L L

SO

L

KO n

SO KO

L

n L

EI z x EIE z x

E mv m x z x

E

EI x x dx

m x x dx

E

ψ

ω

ω

ψ

ψ

ψ

ψ

= =

= =

== →

∫

∫

∫ ∫

∫S M Shiyekar, Professor Civil Department

2

2 0

2

0

( )[ ''( )]

: Rayleigh Quotient

( )[ ( )]

L

n L

EI x x dx

m x x dx

ψ

ω

ψ

=

∫

∫

Continuum System: Cantilever Beam by Rayleigh-Ritz Method

( ) ( )N

ju x Z xψ=∑1

( ) ( )

Gen. Co-ord., ( ) Shape Function

Selection of Shape Function

1. Must be Continuous

2. Also derivative must be continuous

j

j

j

u x Z x

Z x

ψ

ψ

=

=

= =

∑


Find Natural Frequencies of Cantilever Beam by Rayleigh-Ritz Method

1

1( ) 1 cos

2

xx

L

πψ

= −

2

3( ) 1 cos

2

xx

L

πψ

= −


1 2

" " " "

11 1 1 12 1 2

0 0

" " " "

1 3( ) 1 cos ; ( ) 1 cos

2 2

( )[ ( ) ( )] ; ( )[ ( ) ( )] ;

( )[ ( ) ( )] ; ( )[ ( ) ( )] .

L L

L L

x xx x

L L

K EI x x x dx K EI x x x dx


π πψ ψ

ψ ψ ψ ψ

ψ ψ ψ ψ

= − = −

= =

= =

∫ ∫

∫ ∫


" " " "

21 2 1 22 2 2

0 0

11 1 1 12 1

0

( )[ ( ) ( )] ; ( )[ ( ) ( )] .

( )[ ( ) ( )] ; ( )[ (

L


m m x x x dx m m x x

ψ ψ ψ ψ

ψ ψ ψ

= =

= =

∫ ∫

∫ 2

0

21 2 1 22 2 2

0 0

) ( )] ;

( )[ ( ) ( )] ; ( )[ ( ) ( )] .

L

L L

x dx

m m x x x dx m m x x x dx

ψ

ψ ψ ψ ψ= =

∫

∫ ∫S M Shiyekar, Professor Civil Department


{ }{ }

4

3

2

1 0 0.2267 0.5755;

0 81 0.5755 1.924432

0. Solving Eigen Value Problem

24.0059 3.5235

n

EIK m mL

L

K m

EI EI

π

ω φ

ω ω

= =

− =

= =1 22 2

1 22 2

24.0059 3.5235;

:

22.0345 3.5160;

EI EI

L m L m

Exact

EI EI

L m L m

ω ω

ω ω

= =

= =


Disadvantages of Rayleigh-Ritz Method

• Difficult selection of Shape Function

• Suitable for Specific boundary conditions

• Difficult to select additional functions for

accuracyaccuracy

• Difficult to integrate

• Full flooded [k] and [m] matrices

• Not suitable for computer aided numerical

methods.


Finite Element Method

Beam Element

1 1

2 2

1:

1:

u

u

ψ

ψ

=

=2 2

3 3

4 4

1:

1:

1:

u

u

u

ψ

ψ

ψ

=

=

=


Polynomial Function for DOF2 3

1 2 3 4

2

2 3 4

1 1

2 2

2 3

3 1 2 3 4

0 2 3

( 0) 0 0 0

( 0) 0 0 0

( )

i

i

j

u a a x a x a x

duu a a x a x

dx

u x a

u x a

u x L a a L a L a L

= + + +

= = + + +

= = + + +

= = + + +

= = + + +3 1 2 3 4

23

4 2 3 4

( )

( ) 0 2 3

u x L a a L a L a L

duu x L a a L a L

dx

= = + + +

= = = + + +

1 1

2 2

2 3

3 3

2

4 4

1 0 0 0

0 1 0 0

1

0 1 2 3

u a

u a

u aL L L

L Lu a

=


Polynomial Function for DOF

2 3

1 2 3 4

2

2 3 40 2 3

i

i

j

u a a x a x a x

duu a a x a x

dx

= + + +

= = + + +

1 0 0 0 1 1

2 2

2 3

3 3

2

4 4

1 0 0 0

0 1 0 0

1

0 1 2 3

u a

u a

u aL L L

L Lu a

=

1 1

2 2

2 23 3

4 4

3 2 3 2

1 0 0 0

0 1 0 0

3 2 3 1

2 1 2 1

a u

a u

a uL L L L

a u

L L L L

= − − −


{ }

{ }

1

22 3

3

4

1

22 3

1

1 0 0 0

0 1 0 0

3 2 3 1

i

a

au x x x

a

a

u

u

=

=

Shape Function

{ } 22 3

2 23

4

3 2 3 2

2 2 2 3

1 22 3 2

2 3 2 3

3 42 3 2 2

3 2 3 11

2 1 2 1

3 2 21 ; ;

32 ;

ux x x

uL L L L

u

L L L L

x x x xx

L L L L

x x x x

L L L L

ψ ψ

ψ ψ

= − − −

= − + = − +

= − = − +


Element Stiffness and Mass Matrices

2 2 2 3

1 22 3 2

2 3 2 3

3 42 3 2 2

3 2 21 ; ;

32 ;

x x x xx

L L L L

x x x x

L L L L

ψ ψ

ψ ψ

= − + = − +

= − = − +

" "

0

0

( )[ ( ) ( )] ,

( )[ ( ) ( )]

L

ij i j

L

ij i j

K EI x x x dx

m m x x x dx

ψ ψ

ψ ψ

=

=

∫

∫

L L L L


Element Stiffness and Mass Matrices

2 2

3

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

e

L L

L L L LEIK

L LL

L L L L

− − = − − −

−

2 2

2 2

156 22 54 13

22 4 13 3

54 13 156 22420

13 3 22 4

e

L L

L L L LmLm

L L

L L L L

− − = − − − −


Cantilever Beam: Finite Element Method

( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

2 2

1,2

3

2 2

12 6 12 62 2

6 4 6 22 2 2 2

12 6 12 62 22

6 2 6 42 2 2 2

e

L L

L L L LEI

K

L LL

L L L L

−

− =

− − −

− S M Shiyekar, Professor Civil Department


[ ] [ ]{ }{ }20

nK m ω φ − =

Solution of Eigen Value ProblemSolution of Eigen Value Problem

• Power Method

• Stodola Method


Cantilever Beam: Finite Element Method, Results

No. of Finite Elements

Mode 1 2 3 4 5 Exact

1 3.5327 3.51772 3.51637 3.51613 3.51606 3.51602

2 34.8069 22.2215 22.1069 22.0602 22.0455 22.03452

3 75.1571 62.4659 62.1749 61.9188 61.6972

4 218.138 140.671 122.657 122.320 120.902

5 264.743 228.137 203.020 199.860

6 527.796 366.390 337.273 298.556

Ref: Dynamics of Structures, Theory and Applications to Earthquake Engineering, II nd Edition,

Anil K Chopra, PHI Publications, New Delhi 2005 S M Shiyekar, Professor Civil Department

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finite element and numerical methods in structural dynamics · ref: dynamics of structures, theory...

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