Dynamic Analysis Using FEM

8/19/2019 Dynamic Analysis Using FEM

1/47

ANALYSIS USING FEMANALYSIS USING FEM

iman asu


2/47

Text Book “D namics of Structures” b

Prof A K Chopra Text Book “D namics of Structures” b

Prof J L Humar

“

Analysis” by T Y Yang

taken from the course materials prepared

,


3/47

REVIEW OF DYNAMICS AND

SEISMIC EXCITATION


4/47

Single Degree of Freedom (SDOF) System• One variable is required to describe the deformation• For example: Mass-Spring-Dashpot System

Newton’s Law of Motion• Resultant force

• Resultant Force t ,

Force offered by spring,

Force resisted by dashpot,

kx

cx


5/47

Equation of Motion

• Second order, Non-homogeneous, Ordinary

( )mx cx kx f t + + =

•

Initial Conditions:

Provided system parameters satisfy

certain condition


6/47

Multi-De rees of Freedom

5-Storied Building with

• Columns axially inextensible

• Beams are rigid

5 independent displacements are

required to specify the lateral

deformation profile

5 DOFs, one at each floor

5 (linearly) independent

Mode-1 Mode-2 Mode-3 Mode-4 Mode-5

12345

12345

12345

0

1

2

3

4

01234

0 0.25 0.5 0.75 1 -1.5 -1 -0.5 0 0.5 1 -1.5 -1 -0.5 0 0.5 1 -2-1.5-1 -0.5 0 0.5 1 -4 -3 -2 -1 0 1 2 3

Any instantaneous deformation profile is a linear combination of theseshapes


7/47

with a mass- spring-dashpot

assembly

• Floor mass contributes to the

mass

• Columns contribute to the

spring stiffness

• Dashpot accounts for theinherent damping

e equ va ence o an gures — pr ngs

and dashpots are acting laterally


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Dynamic Equilibrium of MDOF

[ ]{ } [ ]{ } [ ]{ } { }u C u K u P + + = System

[ ]

[ ]

[ ]

Mass Matrix (Diagonal here)

Damping Matrix (Tri-diagonal here)

Stiffness Matrix (Tri-diagonal here)

M

C

K

=

=

=

{ } Excitation Vector P =

[M] (and [C]) has the similar

Force EquilibriumApplied Force=Resistive Force

i


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Under Seismic Excitation

Spatially Uniform

Ground Motion

[ ]{ } [ ]{ } [ ]{ } { }

{ } [ ]{ } g

u C u K u P

P M u

+ + =

= −


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0.1

0.2

a t i o n ( g )

0 10 20 30 40Time sec

-0.2

-0.1 A c c e l e

0

0.1

.

c e l e r

a t i o n ( g )

NS

0 10 20 30 40Time (sec)

-0.2

- . A

0.2

-0.1

0

0.1

c c e l e r a t i o n ( g )

Vertical

0 10 20 30 40

Time (sec)

-0.2


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( ) ( ) g mx cx kx f t mu t + + = = − Response Spectrum

( )22

( , )

n n g

n

x x x u t

S MaxR T

ξω ω

ξ

+ + = −

=

0.4

.

l e r a t i o n

( g )

0.2

p e c t r a l

a c c

0 2 4 6Period (sec)

0

0.4

.

l e r a t i o n

( g )

0.2

p e c t r a l a c c

e

Vertical

0 2 4 6Period (sec)

0


12/47

REVIEW OF ANALYTICAL MECHANICS


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Generalized Coordinate andGeneralized Coordinate and

Constraint EquationsConstraint Equations

1 2, ,...i i N

ii j

y f q q

yq s

=

∂= ∑

j j=

Virtual Displacement

1 1 1

M M N i

i i i j

i i j j

y F y F q

q

= = =

∂=

∂∑ ∑ ∑

Work done = ZERO

1 1 1

Generalized Force

N M N i

i j j j

j i j j

y F q Q q

q

= = =

⎛ ⎞∂ ⎟⎜ ⎟= =⎜ ⎟⎜ ⎟⎜ ∂⎝ ⎠∑ ∑ ∑

1

Equilibrium M

i j i

i j

yQ F zero for

q=

∂= =

∂∑


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L

y q q j

⎛ ⎞⎟⎜= − ⎟⎜

2 1 22

L y q a q j

⎝ ⎠⎛ ⎞⎡ ⎤ ⎟⎜= − −⎢ ⎥ ⎟⎜ ⎟⎜ ⎢ ⎥⎝ ⎠⎣ ⎦

Relation between Generalized and

Constraint Coordinate3 1

4 1 22

L y q q j⎛ ⎞⎟⎜= + ⎟⎜ ⎟⎜⎝ ⎠⎡ ⎤

( )

1 2 1 21

2

1 2

2

22

mg q

L P aq L

k k

−⎪ ⎪+ − − ⎪ ⎪⎧ ⎫⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪ = ⎛ ⎞⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎟⎜⎪ ⎪ ⎪ ⎪−⎢ ⎥ ⎟⎜⎪ ⎪⎩ ⎭ ⎪ ⎪⎟⎜− − ⎝ ⎠⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦

qu r um quat on

obtained by equating

generalized forces to zero

Constraint Equations

( )( )

1 1 2

2 1 2

, ,..., 0, ,..., 0

g y y t g y y t

==

Holonomic constraints


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Exam le of Constraint in Buildin Anal sis

3 Dofs Per Floor Level


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’’In Generalized Coordinate

( )1 2 1 2, ,...., , ,....T T q q q q= Kinetic Energy

1 2, ,.... o en a nergy

Lagrange’s Equation

0 j j j j

d T T V Q

dt q q q

⎛ ⎞∂ ∂ ∂⎟⎜ ⎟− + − =⎜ ⎟⎜ ⎟⎜∂ ∂ ∂⎝ ⎠

jQ = Those forces that cannot be derived from a scalarfunction, e.g., damping force, externally applied forces

etc.


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STATIC ANALYSIS TO DYNAMIC

ANALYSIS


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Potential Energy = Strain EnergyV U =

22

2

02

L

EI v dx x

⎛ ⎞∂ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜∂⎝ ⎠∫

2 L

mass density

= area

=0

Kinetic Energy2

T v x dx= ⎣ ⎦

0 jd T T V

Qdt

⎛ ⎞∂ ∂ ∂⎟⎜ ⎟− + − =⎜ ⎟⎜ ⎟⎜∂ ∂ ∂ Lagrange Equation

( ) ( )2 2

1 2 2

1 1 1 10 0

22

L Ld T V d A v v

Y v x v x dx EI dxdt v v dt v x v x

ρ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎟⎜⎟ ⎟ ⎟⎜ ⎜ ⎜⎟⎜ ⎡ ⎤ ⎡ ⎤ ⎟⎟ ⎜ ⎟ ⎟= + = +⎜ ⎟ ⎜ ⎜⎟⎜⎟ ⎟ ⎟⎜⎟ ⎣ ⎦ ⎣ ⎦⎜ ⎜ ⎜⎟⎜ ⎟ ⎟⎟ ⎜ ⎜⎜ ⎝ ⎠⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠∫ ∫

11 1 12 1 13 2 14 2 11 1 12 1 13 2 14 2m v m m v m K v K K v K θ θ θ θ = + + + + + + +

( ) ( )'' ''

0

L

ij i j K EI f x f x dx⎡ ⎤= ⎢ ⎥⎣ ⎦∫

( ) ( )0

L

ij i jm A f x f x dx ρ ⎡ ⎤= ⎢ ⎥⎣ ⎦∫


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2 2

1 1

2 2

12 6 12 6

156 22 54 13 6 64 2

L L L LY v L L

⎡⎢ −

⎧ ⎫ ⎧ ⎫⎡ ⎤−⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪ −⎪ ⎪ ⎪ ⎪1v

⎤⎥⎢ ⎥

⎢ ⎥⎧ ⎫⎪ ⎪⎢ ⎥⎪ ⎪⎪ ⎪⎢ ⎥⎪ ⎪1 1

2 2

2 22 2

2 2

54 13 156 22 12 6 12 6420

13 3 22 4

6 62 4

Y v L L L

L L L L M L L L L θ

−⎪ ⎪ ⎪ ⎪⎢ ⎥= +⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎢ ⎥ − − −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪− − −⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭−

1

2

2

v

θ

⎪ ⎪⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎪ ⎪⎩ ⎭⎢ ⎥⎢ ⎥

L L⎣ ⎦

Consistent Mass Matrix

• Damping Terms can be added separately

• Element Equilibrium Equations can be assembled to form

Global structural e uilibrium


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USE of FEM in DYNAMIC ANALYSIS


21/47

Divide the continuum into a finite number of sub-

re ions elements of sim le eometr

Select key points on the elements to serve as nodes,where conditions of equilibrium and compatibility are tobe enforced.

- -relationships within a typical element

Determine stiffnesses and equivalent nodal loads for a

typical element using work or energy principles

dx dy dz

Develop equilibrium equations for the node ofdiscretized continuum in terms of the elementcontributions

{ }u⎡ ⎤⎣ ⎦ { }u

displacements Calculate stresses at selected points within the elements

Determine support reactions at restrained nodes ifes re


22/47

Overview of FEM for DynamicOverview of FEM for Dynamic

LoadsLoads Divide the continuum into a finite number of sub-regions of

Select key points on the elements to serve as nodes, whereconditions of equilibrium and compatibility are to be enforced.

- -

a typical element

Determine stiffness, mass, damping and equivalent nodal

dynamic loads for a typical element using work or energyprinciples

Develop dynamic equilibrium equations for the node of

discretized continuum in terms of the element contribution

dx y

o ve t ese equi i rium equations or t e no a isp acements Calculate stresses at selected points within the elements

Determine support reactions at restrained nodes if desired

{ }dV u ρ

{ }cdV u { }u


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Displacement formulation

( , , , ) f f u u x y z t =

Displacement field is function of timedx dy dz

Nodal displacements are functions of time

u u t =

The spatial interpolation is unchanged

{ } [ ]{ }( , , , ) ( , , ) ( ) f u x y z t N x y z u t =


24/47

{ } ρdV u

cdV u

dx dy dz For general volumetric loads

Infinitesimal element in 3D( ). .

T

e n l

V

N cu u dV ρ = − −∫

• e spa a n erpo a on

{ } ( ) ( ){ }, ,u N x y z u t ⎡ ⎤= ⎣ ⎦

. .

. .

T T

e n l

V V

e n l

f cN NdVu N NdVu

f cu mu

ρ = − −

= − −

∫ ∫

{ } ( ) ( ){ }, ,

, ,

u N x y z u t

u N x z u t

⎡ ⎤= ⎣ ⎦

⎡ ⎤=

Apply D’Alembert’s principle

ku t = +. .

( )mu cu ku p t + + =


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Equation of motion for system is ( )mu cu ku p t + + =

dy dzp(t)

Special case is seismic load in 3-Dimension

+ + = − − − mu cu ku m u m u m u

Resultant force vector

= − − − x xg y yg z zg p m u m u m udx

dy dz

ig u

is the ground acceleration in direction i and mi is a column matrix which

represents the sum of all columns in the mass matrix m associated with

ig u

This definition is valid only if the vector is defined as the displacement relative to

the displacement at the base of the structure


26/47

Overview of Dynamic Problems andOverview of Dynamic Problems and

Analysis TypesAnalysis TypesProblems Analysis

Rigid body vs elasticmotion

Real eigen valueanalysis

Eigen value

vibration

Loadin t e

analysis

Fre uenc res onseo Harmonic

o Periodic

Transient response

Random response

Response

o ransiento Stationary random

o random


27/47

Infinite degree of freedom Equation of motion is PDE

x

Finite Element Model

Continuous mass is discretized

u x,

ODE x1

Finite degree of freedom

Assumed displacement field

u(x1 ,t)ODE

qua on o mo on s

Discretization of mass can be done

usin two formulations

x3

u(x1 ,t)

x1 x2

Consistent Mass

Lumped Mass

u(x3 ,t) ,


28/47

General approach for element mass matrix

T

V m N NdV ρ = ∫

Alternativel

1 1T

V

m h g gdVh ρ − −= ∫ where = geometric matrix

g

f u gc

u hc

=

=

1c h u−=

1 f u gh u−=

= un erm ne cons an s

= displacement field

=

f u

u

1 N gh−=

= shape function N

For different types of elements, geometric matrix is different andhence equivalent mass matrix is also different


29/47

For 1D flexure beam element, considering thetranslational inertia only

1 1T m h g gdVh ρ − −= ∫ V

2 31 g x x x⎡ ⎤= ⎣ ⎦

2 3 41 1 1

2 3 4 L L L L

⎡ ⎤⎢ ⎥⎢ ⎥ 3 0 0 0 L⎡ ⎤

2 3 4

2 3 5 6

T

x x x

x x x x g g

x x x x

⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥

2 3 4 5

3 4 5 60

2 3 4 5

1 1 1 1

3 4 5 6

L

T

L L L L

g gdx

L L L L

⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥

⎢ ⎥

∫

3

1

3 2 2

0 0 01

3 2 3

Lh

L L L L L

−

⎢ ⎥⎢ ⎥=⎢ ⎥− −

⎢ ⎥−

x x x x 4 5 6 71 1 1 1

4 5 6 7 L L L L⎢ ⎥

⎣ ⎦

Consistent Element Mass Matrix

−

Beam

2 222 4 13 3

54 13 156 22420

L L L L ALm

L L

ρ ⎢ ⎥−⎢ ⎥=⎢ ⎥−

2 213 3 22 4 L L L L− − −⎣ ⎦


30/47

For 1D flexure beam element, considerin the rotationalinertia only

where1 1T m h g g dVh ρ − −= 2

, 0 1 2 3 x g x x⎡ ⎤=, ,V

0 0 0 0⎡ ⎤ 2 30 0 0 0⎡ ⎤

⎢ ⎥2

, , 2 3

2 3 4

0 1 2 3

0 2 4 6

0 3 6 9

T

x x

x x g g

x x x

x x x

⎢ ⎥=⎢ ⎥

⎢ ⎥

2 3 4, ,

0

4 30

3 2

3 9

L

T

x x g g dx L L L⎢ ⎥= ⎢ ⎥

⎢ ⎥

∫

2 5⎢ ⎥⎣ ⎦

Rotational Mass 36 3 36 3 L L−⎡ ⎤

Beam

Matrix forBeam Element

2 2

2 2

3 4 336 3 36 330

r L L L I m L L L

ρ −⎢ ⎥=⎢ ⎥− − −⎢ ⎥

Element− − −


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1 3 5

6Two beam elements with three

1 2 3

Assembly of stiffness matrix

[ ]

2 2

2 2 2 23

12 6 12 6 0 0

6 4 6 2 0 0

12 6 12 12 6 6 12 6

6 2 6 6 4 4 6 2

L L

L L L L

L L L L EI K

L L L L L L L L L

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥− − + − + −

= ⎢ ⎥

− + + −⎢ ⎥

123

4

Similarly, Assembly of Mass (consistent) matrix

2 20 0 6 2 6 4 L L L L

− − −⎢ ⎥

−⎣ ⎦56

1 2 3 4 5 6

2 2

156 22 54 13 0 0

22 4 13 3 0 0

54 13 156 156 22 22 22 13

L L

L L L L

L L L L L AL ρ

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥+ − + − −

= ⎢ ⎥32

1

2 2

0 0 54 13 156 22

0 0 13 3 22 4

L L

L L L L

− − − + + − −⎢ ⎥⎢ ⎥−⎢ ⎥

− − −⎣ ⎦5 6431 2

6

5


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Exclude rigid body modes (typical for civil engineering problems)

For example, in two element cantilever beam, Node 3 is restrained nodeso DOFs 5 & 6 can be eliminated1 3 5

1

22 2156 22 54 13 L L−⎡ ⎤

Mass matrix for constrained beam

1 2 3

2 4

3

4

1 2 3 4

2 2

54 13 312 0420

13 3 0 8

m L

L L L

ρ −⎢ ⎥=⎢ ⎥⎢ ⎥− −⎣ ⎦

Stiffness matrix for constrained beam

LL

L1

3

2

1

2 2

3

12 6 12 6

6 4 6 2

L L

L L L L EI k

− −⎡ ⎤⎢ ⎥−⎢ ⎥=

431 2

42 26 2 0 8 L L L

− −⎢ ⎥⎣ ⎦


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Reduction of Size for DynamicReduction of Size for Dynamic

AnalysisAnalysisStiffness (static) reduction

RR RT R RR RT R R

TR TT T TR TT T T

m m u u

m m u k k u F

=+ =

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

RR R RT T u u+ = 1

R RR RT T u k k u−= −

1 2

23

4

Hence, from the second equation

∗

where

TT T T =

1 2 3

1 0TT RR TR RR RT

k k k k k ∗ −= − =


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Mass (dynamic) reduction1−

TT T T m T mT =

RR RT

T

T

T I

−= ⎢ ⎥⎣ ⎦

3

1.7143 4.2857

4.2857 13.714TT

EI k

L

∗ −⎡ ⎤= ⎢ ⎥−1

2 2

2 2

4 2 6 6

2 8 6 0 RR RT

L L L

k k L L L EI ⎡ ⎤−⎢ ⎥

⎡ ⎤ ⎢ ⎥=

2

4

1 33

6 6 12 12

6 0 12 24

TR TT k k L L L

L

⎣ ⎦⎢ ⎥− −⎢ ⎥⎣ ⎦

1

3

2 4 1 3

Similarly, applying mass reduction2 2

2 2

4 3 22 13 L L L L⎡ ⎤−⎢ ⎥

21

420 22 13 156 54

13 0 54 312

RR RT

TR TT m m L L

L

− −

=⎢ ⎥ ⎢ ⎥−⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

1

3

. .

0.17566 0.89096TT m AL ρ ∗ = ⎢ ⎥⎣ ⎦

1 3

3

3142


35/47

Vibration of a CantileverVibration of a Cantilever

Free vibration of a uniform cantilever beam

L

4 2

0u u

EI AL x t

ρ + =∂ ∂

( ) ( ), ni t nu x t x eω

φ =

( )n xφ = characteristic function (nth mode)

L=1000mm, A=(20x5)mm2

E = 2.1x105 N/mm 2

( )( )

4

4

4 0

n

n n

d x x

dx

φ β φ − =

4

n β 2 /n L EI ρ ω =

( )2 4/n nl EI mω β =General solution is = natural re uenc o nth mode

( ) 1 2 3 4sinh cosh sinh coshn x c x c x c x c xφ β β β β = + + +

A l in BCs and solvin the characteristic e uation natural fre uencies of

ω1

ω2

ω3

beam can be obtained

12

43.156

EI

mL

⎛ ⎞⎜ ⎟

⎝ ⎠

12

422.03

EI

mL

⎛ ⎞⎜ ⎟

⎝ ⎠

2

461.70

EI

mL

⎛ ⎞⎜ ⎟⎝ ⎠


36/47

Equation of motion for undamped system is

∗ ∗

( ) cosu t t φ ω =

=

2 0k mω φ ⎡ ⎤− =⎣ ⎦

TT

TT

2 21,11 11 12 12

2 2

0ii ik m k m φ ω ω ⎡ ⎤ ⎡ ⎤− − ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ,21 21 22 22 ii i ,

Solving above equation and substituting L = L1/2

1 2

1

3.5222

1

L A

EI

ω ρ

=

1.0000 1.0000⎡ ⎤2 2

1

.

L A ρ 1 2

0.3395 0.6991−⎣ ⎦


37/47

Simpler model for inertia properties for

(1) (2)

Various approaches Tributary masses are lumped at the nodes of

L

m

LL1m2

1

Lumped mass matrix m( 1)/2

m( 2)/2

1

2

0 ... 0 ... 0

0 ... 0 ... 0

... ... ... ... ... ...

m

m⎢ ⎥⎢ ⎥⎢ ⎥

( )( ) ( 1)1

2i i im m m += +

0 0 ... ... 0

... ... ... ... ... ...

im m= ⎢ ⎥⎢ ⎥⎢ ⎥

where nn = number of nodes

mi = tributary masses lumped at node i

0 0 ... 0 ...nn

m⎢ ⎥⎣ ⎦


38/47

Using the same and Lumped Mass Matrix (LMM)

different set of ω and Ф are obtainedTT k ∗

LTT m

∗

mm2

1 0.5 00 1.0

TT m AL ρ ∗ ⎡ ⎤= ⎢ ⎥

1

3

2 21,11 11 12 12 0ii ik m k m φ ω ω ⎡ ⎤ ⎡ ⎤− − ⎡ ⎤

=⎢ ⎥ i = 1 2

L11 3

2,21 21 22 22 ii im mω ω − −

Solving above equation and substituting L = L1/2

1 2

1

13.156 EI L A

ω ρ

=

1.0000 1.0000⎡ ⎤

2 2

1

16.2584 A

ω ρ

= 1 2 0.3273 1.5272= =

−⎣ ⎦


39/47

Lumped Mass MatrixConsistent Mass Matrix

1 21

1

3.156

I

Aω ρ =1 21

1

3.5222

EI

Aω ρ =

2 2

1

16.2584 A

ω ρ

=

1 2

1.0000 1.0000⎡ ⎤Φ = Φ Φ =

2 2

1

22.28 A

ω ρ

=

1 2

1.0000 1.0000⎡ ⎤Φ = Φ Φ = . .−. .−

(0.327)LM

(0.339)CM

ω1

ω2

(-1.53)LM

(1.0)CM,LM

12 EI

12 EI

(-0.699)LM (1.0)CM,LM

4.

mL⎝ ⎠ 4

.mL⎝ ⎠


40/47

Lumped mass matrix (LPM)

L

L=1000mm,

A=(20x5)mm2

• More number of elements –

better solution

E = 2.1x10^5 N/mm2

Consistent mass matrix (CMM)• LPM -lower bound solution,

CMM - upper bound solution

• For same number of element

CMM shows better results


41/47

Proportionally Damped System

Equation of motion can be dynamically uncoupled by the eigenvectors of the

undamped system

0+ + = mu c u ku

α β = + P c m k

Rayleigh Damping

c = αm+βkζr

2 i jω ω α ζ =

2 β ζ =

=

c = βk

i j i j

ζ = damping ratio

ω ω = natural re uenc o i th and th mode Transform to normal coordinates

ωr

[ ] 2T P P r r r C c M φ φ ζ ω = =


42/47

u C u K u P + + =

Governing Equation

{ } [ ]{ } g P M u= −


43/47


44/47

0.15

0.2

-0.05

0

0.05

0.1

0 5 10 15 20 25 30 35 40 e l e r a t i o n

( g )

-0.2

-0.15

-0.1 A c

Time (sec)

Ground Motion Along Y-direction


45/47

In KN


46/47

In mm


47/47

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Documents

Dynamic Analysis Using FEM