Computer Aided Ship Design Part.3 Grillage Analysis of Midship …ocw.snu.ac.kr/sites/default/files/NOTE/6324.pdf · 2018. 1. 30. · 2009 Fall, Computer Aided Ship Design, Part3

SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.krSeoul NationalUniv.

2009 Fall, Computer Aided Ship Design, Part3 Grillage Analysis of Midship Cargo Hold 01- Bar Element

Naval Arc

hitectu

re &

Ocean E

ngin

eering


Computer Aided Ship DesignPart.3 Grillage Analysis of Midship

Cargo Hold

2009 Fall

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering,Seoul National University of College of Engineering



Application

Bar

Beam

Shaft

Summary

Element Behavior

Tension

Bending

Torsion

Midship Cargo Hold

2

2

( )( ) 0

d u xEA f x

dx

4

4

( )( ) 0

w xEI f x

x

Structure

•Superposition of Stiffness Matrix•Coordinates Transformation

Truss

Frame

Grillage2

2

( )( ) 0

xGJ f x

x

Beam Theory : Sign Convention, Deflection of Beam

Elasticity : Displacement, Strain, Stress, Force Equilibrium, Compatibility, Constitutive Equation

:A Sectional Area :

:

E

I

Young’s Modulus

Moment of Inertia

:G Shear Modulus

:J Polar Moment of Inertia:l Length

Differential Equation

Mx F , 0where x

VariationalMethod

2

0( ) 0

2

l EA duf u dx

dx

22

20( ) 0

2

l EA d wf w dx

dx

2

0( ) 0

2

l GJ df dx

dx

1 1

2 2

1 1

1 1

u fEA

u fl

0 1( )u x a a x

11

2 2

11

3

2 2

2 2

2 2

6 3 6 3

3 2 32

6 3 6 3

3 3 2

ful l

Ml l l lEI

ul l fl

l l l l M

2 3

0 1 2 3( )w x b b x b x b x

1 1

2 2

1 1

1 1

MGJ

Ml

0 1( )x c c x

Kd F

Finite Element Method

•Discretization•Approximation

Grillage Modeling

•Equivalent Force & Moment•3D 2D

•Boundary condition

Engineering Concept !

Solution

•programming•visulaization

:u

Vertical Displacement

:G Shear Modulus

: Angle of Twist:w

Axial Displacement

2/41



Chapter 1. Element : Bar

3/41



Element : Bar - Differential Eqn.

Rao,S.S.,Mechanical Vibrations, Fourth Edition, Prentice Hall, 2004,

z

xo

ac

bdx dx

( , )f x t

l

a

b

c

d

f dx

P dP

dx

Equilibrium position a

b

c

d u

u du

Displacedposition

P

( )u x

x

l

( )f x

( )u x

x0 l

an example of displacement in x-direction at x

:density, : Young s Modulus, :sectionalareaE A

( , ):external force per unit lengthf x tif is constantAand is time invariantf

P : the axial forces acting on the cross sections of a small element of the bar of length dx

4/41



Element : Bar - Differential Equ.

A Bar in Axial Vibration

2

2

( , )( ) ( )

u x tP dP f dx P A x dx

t

F ma

2

2

( , )( )

u x tdP f dx A x dx

t

2

2

( , ) ( , )( ) ( , ) ( )

u x t u x tEA x dx f x t dx A x dx

x x t

( , ):external force per unit lengthf x t

2

2

( )( ) 0

d u xEA f x

dx 0

u

t

2 2

2 2

( , ) ( , )( , )

u x t u x tEA f x t A

x t

dynamics (vibration)

( ) : .if A x A const

( , )( ) ( ) ( )

( , )( )

u x tP A x EA x EA x

x

P u x tdP dx EA x dx

x x x

‘constitutive equation’

“Longitudinal Vibration of a Bar or Rod : Rao,S.S.,Mechanical Vibrations, Fourth Edition, Prentice Hall, 2004, pp597-560

subjected to

0

0

( , 0) ( ), 0

( , 0) ( ), 0

u x t u x x l

ux t u x x l

t

I.V.P

(0, ) 0, 0

( , ) 0 or ( , ) 0, 0

u t t

u uAE l t l t t

x x

at the free end , axial force

B.V.P

statics

:density, : Young s Modulus, :sectionalareaE A

z

xo

ac

bdx dx

( , )f x t

l

a

b

c

d

a

b

c

d

f dx

P dP

u

u du

dx

Equilibrium position

Displacedposition

P

( )u x

5/41



Element : Bar - Variational MethodDifferential Equation

Boundary condition

2

2

( )( ) 0

d u xEA f x

dx

2

200

l d uEA f u dx

dx

2

0( ) 0

2

l EA duf u dx

dx

00 , 0

xx l

duu EA

dx

multiply by and integrateu

2

20

l d uEA u f u dx

dx

L.H.S:

0

0

ll d udu du

EA u EA f u dxdx dx dx

0

l du duEA f u dx

dx dx

2

0( )

2

l EA duf u dx

dx

integration by part

d du u

dx dx

21

2u u u

f u fu

21

2

u u u

x x x

( ) ( )b b

a ah x dx h x dx

operation

6/41



Element : Bar - Rayleigh-Ritz methodDifferential Equation

Boundary condition

2

2

( )( ) 0

d u xEA f x

dx

2

0( ) 0

2

l EA duf u dx

dx

00 , 0

xx l

duu EA

dx

Variational Method2

1 2u a x a x

2

2

1 2 1 20

2 02

l EAa a x f a x f a x dx

1

n k

kku a x

Rayleigh-Ritz method

assume,

2

0( ) 0

2

l EA duf u dx

dx

solution

7/41



(solution)

2 3 32 2

1 2 1 1 2 2

4

2 3 3

f l l f lEAl a EAl a a EAl a EA a a

2

2

1 2 1 20

22


2 2 2 21 1 2 2 1 20

4 42

l EAa a a x a x f a x f a x dx

3 2 32 2 2

1 1 2 2 1 2

0

42

2 3 2 3

l

EA x x xa x a a x a f a f a

3 2 32 2 2

1 1 2 2 1 2

42

2 3 2 3

EA l f l f ll a l a a a a a

3 2 32 2

1 1 1 2 1 2 2 2 1 2

82 2 2

2 3 2 3

EA l f l f ll a a l a a l a a a a a a

3 2 32 2

1 1 1 2 1 2 2 2 1 2

4

3 2 3

l f l f lEAl a a EAl a a EAl a a EA a a a a

8/41



(solution)

2 3 32 2

1 2 1 1 2 2

4

2 3 3

f l l f lEAl a EAl a a EAl a EA a a

2

2

1 2 1 20

22


22

1 2

3 32

1 2

02

40

3 3

f lEAl a EAl a

l f lEAl a EA a

2

2

1 2 1 20

2 02


since

22

13

322

24

33

f ll l

aEA l

a f ll

9/41



(solution)

1

2

2

f l

a EA

fa

EA

5 5

1

4 4 42

2

3 3 3

2 3

f l f l

a

EAla f l f l

23

21

4 322

43 2

3

3

f ll

a l

EAla f ll l

5

4 4

3 3

26

f l f l

EA

fEAl f l

EA

22

13

322

24

33

f ll l

aEA l

a f ll

10/41



Element : Bar - Rayleigh-Ritz methodDifferential Equation

Boundary condition

2

2

( )( ) 0

d u xEA f x

dx

2

0( ) 0

2

l EA duf u dx

dx

00 , 0

xx l

duu EA

dx

Variational Method

1

n k

kku a x

2

1 2u a x a x

Rayleigh-Ritz method

2

2

1 2 1 20

2 02


2( )2

f l fu x x x

EA EA

assume,

2

0( ) 0

2

l EA duf u dx

dx

1

2

2

f l

a EA

fa

EA

11/41



Element : Bar - Finite Element Method

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

x

( )u xl

1u 2u( )u x

1 2( )u x c c x assume: 1 2, (0) , ( )u u u l u

discretization

1 element , 2 nodesfinite element method

12/41




1 2( )u x c c x assume:

1 2( )u l c c l

1 2, ( ) 1x x

or u x u ul l

1 2, (0) , ( )u u u l u

1(0)u c 1 1c u

2 12

u uc

l

2 11( )

u uu x u x

l

1u 2u( )u x 2

0( ) 0

2

l EA duf u dx

dx

Variational Method

13/41




1 2( ) 1x x

u x u ul l

1

2

( ) 1 1 udu x

udx l l

( ) ,u x Nd( )du x

dx Bd

1

2

( ) 1ux x

u xul l

differentiation with respect to x

1

2

1 11 , ,

ux xwhere

ul l l l

N B d

1u 2u( )u x2

0( ) 0

2

l EA duf u dx

dx

Variational Method

14/41




1 2( ) 1x x

u x u ul l

( )u x Nd

( )du x

dx Bd

2

0( ) 0

2

l EA duf u dx

dx

0 0

( ) 02

l lT TEA dx f dx

d B Bd Nd

derivation

1u 2u( )u x

1

2

1 11 , ,

ux xwhere

ul l l l

N B d

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

15/41



(derivation)2

0( )

2

l EA duf u dx

dx

T T0 0

( )2

l lT TEA dx f dx

d B Bd d N

T T T0 0

1( )

2

l lTEA dx f dx

d B B d d N

T T1

2

d Kd d F

T T

d Kd d F

T

d Kd F

0

0

2 0

1

1 1

1

1 1

1 1

1 1

1 1

lT

l

l

EA dx

lEA dx

l l

l

EAdx

l

EA

l

K B B

T TT1 1

2 2 d Kd d K d d F

T

0

1

2

( )l

f dx

u

u

F N

d

1

1 1

x x

l l

l l

N

B

T T T1 1

2 2 d Kd d Kd d F

T T d Kd d K d

TK Ksymmetry

0 0

( )2

l lT TEA dx f dx

d B Bd Nd

T T T :f f f scalar Nd Nd d N Nd

16/41



* * * ** *1 2 1 2

1 1 2 2

1 2

( , ) ( , )( ) ( )

f x x f x xx x x x

x x

(cf. Taylor Series for a Function of Two Variables)

* *

1 2 1 2( , ) ( , )f x x f x x

* *( ) ( ) ( )Tf f f x x x d

*1 * *1

*2 2

, ,x x

x x

x x d x x

* * * ** *1 2 1 2

1 1 2 2

1 2

( , ) ( , )( ) ( )

f x x f x xx x x x

x x

** * * *1 11 2 1 2

*

1 2 2 2

( , ) ( , ) x xf x x f x x

x x x x

* *( ) ( )Tf x x x

* *

1 2

*1 1 1

** *2 21 2

2

( , )

( , )

f x x

x x x

x xf x x

x

T

2 * * 2 * * 2 * ** 2 * * * 21 2 1 2 1 2

1 1 1 1 2 2 2 22 2

1 1 2 2

( , ) ( , ) ( , )1( ) 2 ( )( ) ( )

2

f x x f x x f x xx x x x x x x x R

x x x x

Taylor Series for a Function at ),( 21 xxf ),(*

2

*

1 xx

17/41



* * * ** *1 2 1 2

1 1 2 2

1 2

( , ) ( , )( ) ( )

f x x f x xx x x x

x x


* *

1 2 1 2( , ) ( , )f x x f x x

*1 * *1

*2 2

, ,x x

x x

x x d x x

2 * * 2 * * 2 * ** 2 * * * 21 2 1 2 1 2

1 1 1 1 2 2 2 22 2

1 1 2 2

( , ) ( , ) ( , )1( ) 2 ( )( ) ( )

2


x x x x

2 * * 2 * * 2 * ** 2 * * * 21 2 1 2 1 2

1 1 1 1 2 2 2 22 2

1 1 2 2

( , ) ( , ) ( , )1( ) 2 ( )( ) ( )

2

f x x f x x f x xx x x x x x x x

x x x x

2 * * 2 * * 2 * * 2 * ** 2 * * * * * 21 2 1 2 1 2 1 2

1 1 1 1 2 2 1 1 2 2 2 22 2

1 1 2 1 2 2

( , ) ( , ) ( , ) ( , )1( ) ( )( ) ( )( ) ( )

2

f x x f x x f x x f x xx x x x x x x x x x x x

x x x x x x

2 * * 2 * * 2 * * 2 * *

* * * *1 2 1 * *2 1 2 1 21 1 2 21 1 2 2 1 1 2 22 2

1 1 2 1 2 2

( , ) ( , )( ) (

( , ) ( , )1( ) ( ) ( ) ( )

2)

f x x f x x f x x f x xx x x x x x x x

x x x xx x x

xx

x

* * * *

1 1 2 2 1 1 2 2

2 2 2 2

2 2

1 2 1 1 2 2

*

1 1

*

2 2

( ) ( ) ( )2

(1

)x x x x xf f f f

x

x

x x x xx x x

x

x

x x


2

*

1 xx

18/41



* * * ** *1 2 1 2

1 1 2 2

1 2

( , ) ( , )( ) ( )

f x x f x xx x x x

x x


* *

1 2 1 2( , ) ( , )f x x f x x

*1 * *1

*2 2

, ,x x

x x

x x d x x

2 * * 2 * * 2 * ** 2 * * * 21 2 1 2 1 2

1 1 1 1 2 2 2 22 2

1 1 2 2

( , ) ( , ) ( , )1( ) 2 ( )( ) ( )

2


x x x x

2 * * 2 * * 2 * ** 2 * * * 21 2 1 2 1 2

1 1 1 1 2 2 2 22 2

1 1 2 2

( , ) ( , ) ( , )1( ) 2 ( )( ) ( )

2

f x x f x x f x xx x x x x x x x

x x x x

* * * *

1 1 2 2 1 1 2 2

2 2 2 2

2 2

1 2 1 1 2 2

*

1 1

*

2 2

( ) ( ) ( )2

(1

)x x x x xf f f f

x

x

x x x xx x x

x

x

x x

2 2

2 *1 1 2 1 1

*2 22 2

2

2

* *

1

1 2

1 2 2

1

2

f f

x x x x x

x xf f

x x x

x x x x

* * *1( ) ( ) ( ) ( )2

T Tf f f R x x x d d H x d


2

*

1 xx

19/41



* * * ** *1 2 1 2

1 1 2 2

1 2

( , ) ( , )( ) ( )

f x x f x xx x x x

x x


* *

1 2 1 2( , ) ( , )f x x f x x

*1 * *1

*2 2

, ,x x

x x

x x d x x

2 * * 2 * * 2 * ** 2 * * * 21 2 1 2 1 2

1 1 1 1 2 2 2 22 2

1 1 2 2

( , ) ( , ) ( , )1( ) 2 ( )( ) ( )

2


x x x x

2 2

2 *1 1 2 1 1

*2 22 2

2

2 1

* *

1 1 2

2

2

1

2

f f

x x x x x

x xf fx

x

x x

x x

x

** * * *1 11 2 1 2

*

1 2 2 2

( , ) ( , ) x xf x x f x x

x x x x

* * *1( ) ( ) ( ) ( )2

T Tf f f R x x x d d H x d


2

*

1 xx

20/41




1 2( ) 1x x

u x u ul l

( )u x Nd

( )du x

dx Bd

2

0( ) 0

2

l EA duf u dx

dx

0 0

( ) 02

l lT TEA dx f dx

d B Bd Nd

derivation

1u 2u( )u x

1

2

1 11 , ,

ux xwhere

ul l l l

N B d

T

0 d Kd F

-Chapter 1. Element : Bar

1 1,

1 1

EAwhere

l

K

T

0, ( )

l

f dx F N Kd F

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

21/41



x

( )u x


1u 2u( )u x

1

2

1 1

1 1

uEA

ul

F

T

0, ( )

l

f dx F N

Kd F

( ) :f x f const

constant external force per unit lengthequivalent nodal forces

equivalent nodal forces

x

( )u xl1 2( )u x c c x assume:

1 2, (0) , ( )u u u l u

T

0( )

l

f dx N 0

1l

x

lf dx

x

l

2

2

0

2

2

l

xx

lf

x

l

1

2

f

f

F

1u 2u( )u x

1

1

2f f 1

1

2f f

1

2

1

2

f l

f l

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

22/41



x

( )u x


1u 2u( )u x

1

2

1 1

1 1

uEA

ul

F

T

0, ( )

l

f dx F N

Kd F

( ) :f x f const

constant external force per unit length

equivalent nodal forces

x

( )u xl1 2( )u x c c x assume:

1 2, (0) , ( )u u u l u

x

( )u x

( ) :f x f const

boundary conditionequivalent nodal forces free body diagram

( )u x

( ) :f x f const

f l

1u 2u( )u x

1

1

2f f 1

1

2f f

1

2

1

2

f l

f l

1

2

f

f

F

1

2

1

2

f l

f l

1

2

f lf

f

F

2

1

1 1 2

1 1 1

2

0f l

EA

ulf l

and 1 0u

1 0u 2u( )u x

1

1

2f f 1

1

2f f

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

23/41




x

( )u x

( ) :f x f const

1

2f l

free body diagram

f l

1

2f lf l 1

4f lequivalent

nodal forces

f l

1

4f l

f l

( ) :f x f const ( ) :f x f const

2

4f l

Kd F2

1

01 1 2

1 1 1

2

f lEA

ulf l

1 0u 2u( )u x

1

1

2f f

1

1

2f f

1 element , 2 nodes

1 0u 3u( )u x

1

3

4f f

3

1

4f f

2 element , 3 nodes

2u

2

2

4f f

l

2

l

4

l 2

4 4

l

/ 2ll

F.E.M

2

3

3

41 1 0 02

1 1 1 1/ 2 4

0 1 11

4

f l

EAu f l

lu

f l

superposition of stiffness matrix

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

24/41




x

( )u x

( ) :f x f const

Kd F2

1

01 1 2

1 1 1

2

f lEA

ulf l

1 0u 2u( )u x

1

1

2f f

1

1

2f f

1 element , 2 nodes

1 0u 3u( )u x

1

3

4f f

3

1

4f f

2 element , 3 nodes

2u

2

2

4f f

l

F.E.M

2

3

3

41 1 0 02

1 1 1 1/ 2 4

0 1 11

4

f l

EAu f l

lu

f l


2

1 2

10,

2

f lu u

EA

1 2( ) 1x x

u x u ul l

sol)2

2

3

8

f lu

EA

sol)

2

3

4,

8

f lu

EA

,0 x l

1 0,u

givenfind

displacement solution

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

25/41



(solution)1 0u 3u

( )u x

2 element , 3 nodes

2u

1 2

2 3

1 ,0/ 2 / 2 2

( )

2 1 ,/ 2 / 2 2

x x lu u x

l lu x

x x lu u x l

l l

1 1

1 2( )u x c c x

1 2, (0) , ( / 2)u u u l u 2 3, (0) , ( )u u u l u

2 2

1 2( )u x c c x

) 02

li x ) 2

lii x l

1 2, ( ) 1/ 2 / 2

x xor u x u u

l l

1 2( / 2) / 2u l c c l

1

1(0)u c1

1 1c u

2 12

/ 2

u uc

l

2 11( )

/ 2

u uu x u x

l

0x / 2x l x l

2 2

1 2 2( / 2) / 2u l c c l u 2

1 2 32c u u

2 3 22

/ 2

u uc

l

2 2

1 2 3( )u l c c l u

3 22 3( ) 2

/ 2

u uu x u u x

l

2 3, ( ) 2 1/ 2 / 2

x xor u x u u

l l

26/41




x

( )u x

( ) :f x f const

Kd F2

1

01 1 2

1 1 1

2

f lEA

ulf l

1 0u 2u( )u x

1

1

2f f

1

1

2f f

1 element , 2 nodes

1 0u 3u( )u x

1

3

4f f

3

1

4f f

2 element , 3 nodes

2u

2

2

4f f

l

F.E.M

2

3

3

41 1 0 02

1 1 1 1/ 2 4

0 1 11

4

f l

EAu f l

lu

f l


2

1 2

10,

2

f lu u

EA

1 2( ) 1x x

u x u ul l

sol)2

2

3

8

f lu

EA

sol)

2

3

4,

8

f lu

EA

,0 x l

1 0,u

1 2

2 3

1 ,0/ 2 / 2 2

( )

2 1 ,/ 2 / 2 2

x x lu u x

l lu x

x x lu u x l

l l

givenfind

displacement

given : x

find : ( )u x

2

0( ) 0

2

l EA duf u dx

dx

Variational Method

27/41



Element : Bar - ComparisonDifferential Equation

Boundary Condition

2

2

( )( ) 0

d u xEA f x

dx

2

0( ) 0

2

l EA duf u dx

dx

00 , 0

xx l

duu EA

dx

VariationalMethod

Rayleigh-Ritz Method

2( )2

f l fu x x x

EA EA

x

l ( )u x

( ) :f x f const

2 23

2 2 2 2 8

l f l l f l f lu

EA EA EA

for example,

2

2 1

2 2

f l f f lu l l l

EA EA EA


mathematical modelingby using Newton’s second law

1 2( ) 1x x

u x u ul l

1 2

2 3

1 ,0/ 2 / 2 2

( )

2 1 ,/ 2 / 2 2

x x lu u x

l lu x

x x lu u x l

l l

1 0u 2u( )u x

1

1

2f f

1

1

2f f

1 element , 2 nodes

1 0u 3u( )u x

1

3

4f f

3

1

4f f

2 element , 3 nodes

2u

2

2

4f f

2

1 2

10,

2

f lu u

EA

2 2

1 2 3

3 40, ,

8 8

f l f lu u u

EA EA

2 2/ 2 1 1

2 2 4

l l f l f lu

l EA EA

2

2

1

2

f lu l u

EA

2

2

3

2 8

l f lu u

EA

2 2

3

4 1

8 2

f l f lu l u

EA EA

28/41



Classification

ref. Logan D.L., A First Course in the Finite Element Method, Third edition, Brooks/Cole, p.116

We developed the bar finite element equations by the direct method in Section 3.1 and by the potential energy

method (one or a number of variational methods) in Section 3.10.

In fields other than structural/solid mechanics, it is quite probable that a variational principle, analogous to the

principle or minimum potential energy, for instance, may not be known or even exist. In some flow problems in

fluid mechanics and in mass transport problems (Chapter 13), we often have only the differential equation and

boundary conditions available. However, the finite element method can still be applied.

The methods or weighted residuals applied directly to the differential equation can be used to develop the finite

element equations. In this section, we describe Galerkin's residual method in general and then apply it to the bar

element. This development provides the basis for later applications of Galerkin's method to the nonstructural

heat-transfer element (specifically, the one-dimensional combined conduction, convection, and mass transport

element described in Chapter 13). Because of the mass transport phenomena, the variational formulation is not

known (or certainly is difficult to obtain), so Galerkin's method is necessarily applied to develop the finite element

equations.

Virtual Work

Potential Energy Method

Weighted Residual

Differential Equation Variational Method sol) Rayleigh-Ritz

applicable only to structural problem

sol) F.E.M

Collocation

Least Square

Galerkin

Integral Form

discretize

Algebraic Equation

29/41



-Chapter 1. Element : Bar

Galerkin’s Residual Method

V

R dV

1

2

1 ,ux x

whereul l

N d

Differential Equation2

2

( )0

d u xEA

dx

( )u x Nd

test function

since it is approximated solution

2

2

( )d u xEA

dx 0 Rresidual

Thus substituting the approximated solution into

the differential equation results in a residual over

the whole region of the problem as follows

In the residual method, we require that a weighted value of

the residual be a minimum over the whole region. The

weighting functions allow the weighted integral of

residuals to go to zero

0V

R W dV weighting function or

the basis functions are chosen to play the role of

the weighting functions W

Galerkin Method

1x x

l l

N

1N 2N

iN

0 ,( 1,2)iV

R N dV i

1

0.) ( ) 0ref u v uv xv dx

basis function

30/41



Element : Bar - Galerkin’s Residual Method

the test functions are chosen to play the role of


Galerkin Method

iN

0 ,( 1,2)iV

R N dV i weighting function

residual (test function used)N


2

2

( )0

d u xEA

dx

2

20

( )0 ,( 1,2)

l

i

d u xAE N dx i

dx

Bar - Galerkin’s Residual Method

integration by parts

00

0

ll

ii

dNdu duN AE AE dx

dx dx dx

1 1 2 2, ( )where u x N u N u Nd 1 2, 1 ,x x

N Nl l

1

2

1 1 u

ul l

1

002

1 1,( 1,2)

ll

ii

udN duAE dx N AE i

udx l l dx

1 21 2

dN dNduu u

dx dx dx since

111

002

1 1l

l udN duAE dx N AE

udx l l dx

122

002

1 1l

l udN duAE dx N AE

udx l l dx

1:i

2 :i

1 1

0x l x

du duN AE N AE

dx dx

2 2

0x l x

du duN AE N AE

dx dx

1 2(0) 1, ( ) 0N N l

2 2(0) 0, ( ) 1N N l

sincedu

AE AE A fdx

since

1 0xN f

2 x lN f

1f

2f

1


31/41



Element : Bar - Galerkin’s Residual Method

the test functions are chosen to play the role of


Galerkin Method

iN

0 ,( 1,2)iV

R N dV i weighting function

residual (test function used)N


2

2

( )0

d u xEA

dx

2

20

( )0 ,( 1,2)

l

i

d u xAE N dx i

dx

Bar - Galerkin’s Residual Method

integration by parts

1 1 2 2, ( )where u x N u N u Nd 1 2, 1 ,x x

N Nl l

111

02

1 1l udNAE dx f

udx l l

122

02

1 1l udNAE dx f

udx l l

1 1 1 12 20 02 2 2 2

1 1 1 1 11 1 1 1

l lu u u uAEAE dx AE dx AE l l

u u u ul l l l l l

1 1 1 12 20 02 2 2 2

1 1 1 1 11 1 1 1

l lu u u uAEAE dx AE dx AE l l

u u u ul l l l l l

1 1

2

1 1uAE

ful

1 22

1 1uAE

ful

1 1

2 2

1 1, , ,

1 1

u fEAwhere

u fl

K d F Kd F

1


32/41



Element : Bar - Potential Energy Approach

the principle of minimum potential energy

Of all the geometrically possible shapes that a body can assume,the true one, corresponding to the safisfaction of stable equilibrium of the body, is identified by a minimum value of the total potential energy

the total potential energy Π is defined as the

sum of the internal strain energy Πin and the potential energy of the external forces Πext

in ext

33/41



Element : Bar - Potential Energy Approach the total potential energy Π is defined as the


in ext

1

2in in x x

V V

d dV the strain energy for one-dimensional stress.

x

x

Linear-elastic

(Hooke’s law)material

x xE

0

x

in xd d dxdydz

xd

0 x

0

x

x xE d dxdydz

21

2x xE d dxdydz

1

2xdxdydz

To evaluate the strain energy for a bar,

we consider only the work done by the internal forces during

deformation.

34/41



Element : Bar - Potential Energy Approach

the total potential energy Π is defined as the


in ext

The potential energy of the external forces, being opposite in sign from the extenal work

expression because the potential energy of external forces is lost when the work is done by

the external forces, is given by

11

M

ext b x s ix i

iV S

X udV T u dS f u

l

1xf2xf

bX

xT1S

x

ubXbody forces typically from the self-weight of the bar (in units of force per unit volume) moving

through displacement function

susurface loading or traction typically from distributed loading acting along the surface of

the element (in units of force per unit surface area) moving through displacements

where are the displacements occurring over surface

xT

su 1S

nodal concentrated force moving through nodal displacements iuixf

35/41



Element : Bar - Potential Energy Approach the principle of minimum potential energy


the total potential energy Π is defined as the sum of the internal strain energy Πin and the potential energy of the

external forces Πext

in ext 1

2in x x

V

dV 1

1

,M

ext b x s ix i

iV S

X udV T u dS f u

1. Formulate an expression for the total potential energy.

2. Assume the displacement pattern to vary with a finite set of undetermined parameters (here

these are the nodal displacements ), which are substituted into the expression for total

potential energy.

3. Obtain a set of simultaneous equations minimizing the total potential energy with respect to

these nodal parameters. These resulting equations represent the element equations.

Apply the following steps when using the principle of minimum potential energy

to derive the finite element equations.

iu

36/41







in ext 1

2in x x

V

dV 1

1

,M

ext b x s ix i

iV S

X udV T u dS f u

l

1xf2xf x

1 1 2 202

l

x x x x

Adx f u f u

1. Formulate an expression for the total potential energy.

Apply the following steps when using the principle of minimum potential

energy to derive the finite element equations.dV Adx

2. Assume the displacement pattern to vary with a finite set of undetermined parameters (here these are the nodal

displacements ) ,which are substituted into the expression for total potential energy.

we have the axial displacement function expressed in terms of the shape functions and nodal displacements by

1

2

, 1 ,ux x

whereul l

N d( )u x Nd

assume that there is no surface traction and body force and the

sectional area is constantA

iu

1u2u

37/41







in ext 1

2in x x

V

dV 1

1

,M

ext b x s ix i

iV S

X udV T u dS f u

1 1 2 202

l

x x x x

Adx f u f u

1

2

, 1 ,ux x

whereul l

N d( )u x Nd



1

2

1 1x

udu

udx l l

x xE E Bd

Bd

T T

02

lAE dx Bd Bd d F

1 T1 1 2 2 1 22

x

x x

x

ff u f u u u

f

d F

T T T

02

LEAdx d B Bd d F

l

1xf2xf x

dV Adx1u2u

38/41







in ext 1

2in x x

V

dV 1

1

,M

ext b x s ix i

iV S

X udV T u dS f u

1 1 2 202

l

x x x x

Adx f u f u



1

2

, 1 ,ux x

whereul l

N d( )u x Nd

1

2

1 1x

udu

udx l l

x xE E Bd

Bd

1 T1 1 2 2 1 22

x

x x

x

ff u f u u u

f

d F

T T T

02

lEAdx d B Bd d F

3. Obtain a set of simultaneous equations minimizing the total potential

energy with respect to these nodal parameters. These resulting

equations represent the element equations.

The minimization of Π with respect to each nodal displacement requires that

1

0u

and

2

0u

l

1xf2xf x

dV Adx1u2u

39/41







in ext 1

2in x x

V

dV 1

1

,M

ext b x s ix i

iV S

X udV T u dS f u

1

2

1 1 u

ul l

Bd

1T 1 22

1 1 2 2

x

x

x x

fu u

f

f u f u

d F

T T T

02

lEAdx d B Bd d F

The minimization of Π with respect to each nodal displacement requires that

1

0u

and

2

0u

11 22

1

1 1

1

T T Tul

u uul l

l

d B D Bd 1

1 22

2

2 2

1 1 2 22

1 11

1 1

1( 2 )

uu u

ul

u u u ul

2 21 1 2 2 1 1 2 22 0

( 2 )2

l

x x

EAu u u u dx f u f u

l

1 2 1 1 2 12 01

(2 2 ) ( )2

l

x x

EA EAu u dx f u u f

u l l

1 2 2 1 2 22 02

( 2 2 ) ( )2

l

x x

EA EAu u dx f u u f

u l l

1 2 1

1 2 2

( ) 0

( ) 0

x

x

EAu u f

l

EAu u f

l

In matrix form, we express

11

2 2

1 1

1 1

x

x

fuEA

u fl

11

22

1 1, , ,

1 1

x

x

fuEAwhere

ful

K d F Kd F

40/41



Element : Bar

Bar

Element

2

2

( )( ) 0

d u xEA f x

dx


Mx F , 0where x

VariationalMethod

2

0( ) 0

2

l EA duf u dx

dx

1 1

2 2

1 1

1 1

u fEA

u fl

0 1( )u x a a x

Kd F


•Discretization•Approximation

f1 f2

kNode1 Node 2

δ1 δ2

! Notation

2

1

2

1

kk

kk

f

f

]][[][ Kf

EAk

l

: , :E Young s Modulus A sectional area

stiffness matrix

Galerkin’sWeighted Residual

Energy Method

from now on,we will use this!

41/41

Computer Aided Ship Design Part.3 Grillage Analysis of Midship …ocw.snu.ac.kr/sites/default/files/NOTE/6324.pdf · 2018. 1. 30. · 2009 Fall, Computer Aided Ship Design, Part3

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