Computer Aided Ship Design Part.3 Grillage Analysis of ...

2009 Fall, Computer Aided Ship Design, Part 3. Grillage 02-Beam Element

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

Seoul NationalUniv.

Naval Arc

hitectu

re &

Ocean E

ngin

eering


Seoul NationalUniv.

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



Seoul NationalUniv.

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

2/37



Seoul NationalUniv.

Chapter 2. Element : Beam

3/37



Seoul NationalUniv.

Element : Beam - Differential Eqn.

x ε i

① strain at in x-direction : y

,x x σ i ε i , , y θ k y j E

ds

d ds

1d

ds

( )y d d d yy

ds ds

, : initial length

, :elongated length

ds

y d

ds

x

y

neutralsurface y

dx

d

ds

xσ* neutral surface : Longitudinal lines on the lower part of the beam are elongated, whereas those on the upper part are shortened. Thus the lower part of the beam is in tension and the upper part is in compression. Somewhere between the top and bottom of the beam is a surface in which longitudinal lines do not change in length. This surface is called neutral surface

4/37



Seoul NationalUniv.


x ε i

,x

yE where

σ i i

( )x xd dA dA dA F σ i i③ force acting on in x-direction : dA

④ moment about z-axis :

2

( ) ( )y y

d d y E dA E dA

M y F j i k

② stress at in x-direction :y

① strain at in x-direction : y

,x x σ i ε i , , y θ k y j E

ds

d ds

1d

ds

x

yE

σ i i

yd E dA

F i

( )y d d d yy

ds ds

, : initial length

, :elongated length

ds

y d

neutralsurface

y

dx

d

ds

ds

x

y

xσ

5/37



Seoul NationalUniv.


ds

d ds

1d

ds

① strain at in x-direction :

x ε i

,x x σ i ε i

④ moment about z-axis :

2

( ) ( )y y

d d y E dA E dA

M y F j i k

y

, , y θ k y j

( )y d d dy

ds ds

E

② stress at in x-direction :y x

yE

σ i i

③ force acting on in x-direction : dAy

d E dA

F i

2

A A

yd E dA

M M k

⑤ assume dx

dydxds tan,

2

2

d d y

ds dx

2

2

d d dy d dy d y

ds ds dx dx dx dx

AdAyI 2Define then, ,

EI EIM

M k EI

M k

, : initiallength, :elongatedlengthd y d

2

2,

d yM EI

dx

2

2

d yEI

dxM k

dEI

ds

M k

neutralsurface

y

dx

d

ds

ds

x

y

xσ

6/37



Seoul NationalUniv.

Element : Beam - Differential Eqn.1d

ds


x ε i

,x x σ i ε i

④ moment about z-axis :2

( ) ( )y y

d d y E dA E dA

M y F j i k

y

, , y θ k y j

( )y d d dy

ds ds

E


yE

σ i i


d E dA

F i

2

A A

yd E dA

M M k

⑤ assume dx

dydxds tan,

2

2

d d y

ds dx

2,A

I y dA EI

k


2

2

d yEI

dxM k

x

y

( )f x

neutralsurface

y

dx

d

ds

ds

x

y

xσ

EI dEI

ds

M k k

⑥ relationships between loads, shear forces, and bending moments

1 2 1 2, , ,V M

V V dx M M dxx x

V j V j M k M k

•force equilibrium1 2 ( ) 0y x F V V f

11 1

11 1

( ) 0

( ) 0

VV V dx f x dx

x

VV V dx f x dx

x

j j j

j ( )dV

f xdx

2

2,

d yM EI

dx

7/37



Seoul NationalUniv.


ds


x ε i

,x x σ i ε i


( ) ( )y y

d d y E dA E dA

M y F j i k

y

, , y θ k y j

( )y d d dy

ds ds

E


yE

σ i i


d E dA

F i

2

A A

yd E dA

M M k

⑤ assume dx

dydxds tan,

2

2

d d y

ds dx

2,A

I y dA EI

k


x

y

( )f x

neutralsurface

y

dx

d

ds

ds

x

y

xσ


1 2 1 2, , ,V M

V V dx M M dxx x

V j V j M k M k

•force equilibrium ( )dV

f xdx

( )dM

V xdx

•moment equilibrium 1 2 2

1( ) 0

2z d d x dx M M M x V x f

1

( ) 02

M VM M dx dx V dx dx f x dx

x x

k k i j i j

2

2

d yEI

dxM k

EI dEI

ds

M k k

2

2,

d yM EI

dx

8/37



Seoul NationalUniv.


ds


x ε i

,x x σ i ε i


( ) ( )y y

d d y E dA E dA

M y F j i k

y

, , y θ k y j

( )y d d dy

ds ds

E


yE

σ i i


d E dA

F i

2

A A

yd E dA

M M k

⑤ assume dx

dydxds tan,

2

2

d d y

ds dx

2,A

I y dA EI

k


x

y

( )f x

neutralsurface

y

dx

d

ds

ds

x

y

xσ


1 2 1 2, , ,V M

V V dx M M dxx x

V j V j M k M k

•force equilibrium ( )dV

f xdx

( )dM

V xdx

•moment equilibrium

2

2

d yEI

dxM k

EI dEI

ds

M k k

2

2,

d yM EI

dx

2

2

d y M

dx EI

3

3

1 1( )

d y dMV x

dx EI dx EI

4

4

1 1( )

d y dVf x

dx EI dx EI

4

4( )

d yEI f x

dx

9/37



Seoul NationalUniv.

Element : Beam - Variational MethodDifferential Equation

Boundary condition

4

4( ) 0

d vEI f x

dx

4

400

l d vEI f v dx

dx

multiply by and integrateu

4

40

l d vEI v f v dx

dx

L.H.S:

3 3

3 30 00

ll ld vd v d v

EI v EI dx f v dxdx dx dx

3

30 0

l ld vd vEI dx f v dx

dx dx

2 2 2

2 2 20 00

ll ld v d v d v d v

EI EI dx f v dxdx dx dx dx

integration by part

d dv v

dx dx

21

2v v v

f v fv

( ) ( )b b

a ah x dx h x dx

operation

0

2 2

2 2

0

0 , 0

, 0, 0

x x l

x x l

v v

d v d vEI EI

dx dx

2 2

2 20 0

l ld v d vEI dx f v dx

dx dx

22 2 2

2 2 2

1

2

v v v

x x x

10/37



Seoul NationalUniv.

Element : Beam - Variational MethodDifferential Equation

Boundary Condition

4

4( ) 0

d vEI f x

dx

4

400

l d vEI f v dx

dx

multiply by and integrateu

4

40

l d vEI v f v dx

dx

L.H.S:

integration by part

d dv v

dx dx

21

2v v v

f v fv

22 2 2

2 2 2

1

2

v v v

x x x

( ) ( )b b

a ah x dx h x dx

operation

0

2 2

2 2

0

0 , 0

, 0, 0

x x l

x x l

v v

d v d vEI EI

dx dx

2 2

2 20 0

l ld v d vEI dx f v dx

dx dx

2 2

2 20 0

1

2

l ld v d vEI dx fv dx

dx dx

2

20 2

l EI d vfv dx

dx

11/37



Seoul NationalUniv.

Element : Beam - Finite Element Method

Variational Method

x

( )v xl

1v 2v( )v x

12

2 3

0 1 2 3( )v x c c x c x c x assume: 1 2, (0) , ( )v v v l v

1 2, (0) , ( )dv dv

ldx dx

discretization

1 element , 2 nodesfinite element method

2

2

20 2

l EI d vfv dx

dx

12/37



Seoul NationalUniv.


2 3

0 1 2 3 2( )v l c c l c l c l v

0(0)v c0 1c v

2 1 2 1 22

3 12c v v

l l

Variational Method1v 2v

( )v x

12

22

20 2

l EI d vfv dx

dx

2 3


1 2, (0) , ( )dv dv

ldx dx

1(0)dv

cdx

2

1 2 3 2( ) 2 3dv

l c c l c ldx

1 1c 3 1 2 1 23 2

2 1c v v

l l

13/37



Seoul NationalUniv.


0 1c v 2 1 2 1 22

3 1, 2c v v

l l


( )v x

12

22

20 2

l EI d vfv dx

dx

2 3


1 2, (0) , ( )dv dv

ldx dx

1 1,c

3 1 2 1 23 2

2 1c v v

l l

2 3

1 1 1 2 1 2 1 2 1 22 3 2

3 1 2 1( ) 2v x v x v v x v v x

l l l l

3 2 3 3 2 2 3 3 2 3 2 2

1 1 2 23 3 3 3

1 1 1 1( ) (2 3 ) ( 2 ) ( 2 3 ) ( )or v x x x l l v x l x l xl x x l v x l x l

l l l l

14/37



Seoul NationalUniv.


1

1

1 2 3 4

2

2

( )

v

v x N N N Nv


( )v x

12

22

20 2

l EI d vfv dx

dx

3 2 3 3 2 2 3 3 2 3 2 2

1 1 2 23 3 3 3

1 1 1 1( ) (2 3 ) ( 2 ) ( 2 3 ) ( )v x x x l l v x l x l xl x x l v x l x l

l l l l

3 2 3

1 3

1(2 3 )N x x l l

l

3 2 2 3

2 3

1( 2 )N x l x l xl

l

3 2

3 3

1( 2 3 )N x x l

l

3 2 2

4 3

1( )N x l x l

l

15/37



Seoul NationalUniv.


1

21

1 2 3 42

2

2

( )

v

d v xB B B B

vdx

( ) ,v x Nd

2

2

( )d v x

dx Bd

1

1

1 2 3 4

2

2

( )

v

v x N N N Nv

differentiation with respect to twice x


( )v x

12

22

20 2

l EI d vfv dx

dx

3 2 3 3 2 2 3 3 2 3 2 2

1 1 2 23 3 3 3

1 1 1 1( ) (2 3 ) ( 2 ) ( 2 3 ) ( )v x x x l l v x l x l xl x x l v x l x l

l l l l

1 3

1(12 6 )B x l

l

2

2 3

1(6 4 )B xl l

l

3 3

1( 12 6 )B x l

l 2

4 3

1(6 2 )B xl l

l

3 2 3

1 3

1(2 3 )N x x l l

l

3 2 2 3

2 3

1( 2 )N x l x l xl

l

3 2

3 3

1( 2 3 )N x x l

l 3 2 2

4 3

1( )N x l x l

l

16/37



Seoul NationalUniv.


( )v x Nd

2

2

( )d v x

dx Bd

22

20( ) 0

2

l EI d vfv dx

dx

0 0

( ) 02

l lT TEI

dx f dx

d B Bd Nd

derivation

1

1

1 2 3 4 1 2 3 4

2

2

, ,

v

where N N N N B B B Bv

N B d

Variational Method1v 2v( )v x

12

2

2

20 2

l EI d vfv dx

dx

2 2

3 3 3 3

1 1 1 1(12 6 ) (6 4 ) ( 12 6 ) (6 2 )x l xl l x l xl l

l l l l

T B B

3

2

3

3

2

3

1(12 6 )

1(6 4 )

1( 12 6 )

1(6 2 )

x ll

xl ll

x ll

xl ll

17/37



Seoul NationalUniv.

(derivation)2

2

20( )

2

l EA d vfv dx

dx

T T

0 0( )

2

l lT TEI

dx f dx

d B Bd d N

T T T

0 0

1( )

2

l lTEI 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

lTEI dx K B B

T TT1 1

2 2 d Kd d K d d F

1 1 2 2[ ]Td v v

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 TEI

dx f dx

d B Bd Nd

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

2 2

3

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

l l

l l l lEI

l ll

l l l l

T

0( )

l

f dx F N

18/37



Seoul NationalUniv.

(derivation)

2 2

3 3 3 3 3 3 3 3

2 2 2 2 2 2

3 3 3 3 3 3 3 3

3

1 1 1 1 1 1 1 1(12 6 ) (12 6 ) (12 6 ) (6 4 ) (12 6 ) ( 12 6 ) (12 6 ) (6 2 )

1 1 1 1 1 1 1 1(6 4 ) (12 6 ) (6 4 ) (6 4 ) (6 4 ) ( 12 6 ) (6 4 ) (6 2 )

1( 12 6

x l x l x l xl l x l x l x l xl ll l l l l l l l

xl l x l xl l xl l xl l x l xl l xl ll l l l l l l l

EI

x ll

2 2

3 3 3 3 3 3 3

2 2 2 2 2 2

3 3 3 3 3 3 3 3

1 1 1 1 1 1 1) (12 6 ) ( 12 6 ) (6 4 ) ( 12 6 ) ( 12 6 ) ( 12 6 ) (6 2 )

1 1 1 1 1 1 1 1(6 2 ) (12 6 ) (6 2 ) (6 4 ) (6 2 ) ( 12 6 ) (6 2 ) (6 2 )

x l x l xl l x l x l x l xl ll l l l l l l

xl l x l xl l xl l xl l x l xl l xl ll l l l l l l l

0

l

dx

2 2

3

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

l l

l l l lEI

l ll

l l l l

2 2 2 2 3 2 2 2 2 3

6 6 6 6

2 2 3 2 2 3 4 2 2 3 2 2 3 4

6 6 6 6

2 2

6 6

1 1 1 1(144 144 36 ) (72 84 24 ) ( 144 144 36 ) (72 60 12 )

1 1 1 1(72 84 24 ) (36 48 16 ) ( 72 84 24 ) (36 36 8 )

1 1( 144 144 36 ) (

x xl l x l xl l x xl l x l xl ll l l l

x l xl l x l xl l x l xl l x l xl ll l l l

EI

x xl ll l

0

2 2 3 2 2 2 2 3

6 6

2 2 3 2 2 3 4 2 2 3 2 2 3 4

6 6 6 3

1 172 84 24 ) (144 144 36 ) ( 72 60 12 )

1 1 1 1(72 60 12 ) (36 36 8 ) ( 72 60 12 ) (36 24 4 )

l

dx

x l xl l x xl l x l xl ll l

x l xl l x l xl l x l xl l x l xl ll l l l

0

lTEI dx K B B

19/37



Seoul NationalUniv.


T

0 d Kd F

-Chapter 2. Element : Beam

T

0, ( )

l

f dx F N Kd F

( )v x Nd

2

2

( )d v x

dx Bd

22

20( ) 0

2

l EI d vfv dx

dx

0 0

( ) 02

l lT TEI

dx f dx

d B Bd Nd

derivation

1

1

1 2 3 4 1 2 3 4

2

2

, ,

v

where N N N N B B B Bv

N B d

Variational Method1v 2v( )v x

12

2

2

20 2

l EI d vfv dx

dx

2 2

3

2 2

12 6 12 6

6 4 6 2,

12 6 12 6

6 2 6 4

l l

l l l lEIwhere

l ll

l l l l

K

20/37



Seoul NationalUniv.


1

2 2

1

3

2

2 2

2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

vl l

l l l lEI

l l vl

l l l l

FT

0, ( )

l

f dx F N

Kd F

constant external force per unit length

equivalent nodal forces


2 3

0 1 2 3( )v x c c x c x c x assume:

1 2

1 2

, (0) , ( ) ,

(0) , ( )

v v v l v

v v l

T

0( )

l

f dx N

3 2 3

3

3 2 2 3

3

03 2

3

3 2 2

3

1(2 3 )

1( 2 )

1( 2 3 )

1( )

l

x x l ll

x l x l xll

f dx

x x ll

x l x ll

43 3

4 23 2 3

3 43

4 32

0

2

2

1 4 3 2

2

4 3

l

xx l xl

x xl x l l

fl x

x l

x xl l

1

1

2

2

f

m

f

m

F

1v2v( )v x

1

2fl

1

2fl

2

2

20 2

l EI d vfv dx

dx

Variational Method

1v 2v( )v x

12

x

( )v xl

x

( )v x

( ) :f x f const

2

2

2

12

2

12

lf

lf

lf

lf

21

12fl

21

12fl

21/37



Seoul NationalUniv.


constant external force per unit length


boundary conditionequivalent nodal forces free body diagram

2

f l

2

2

0

12

0

12

lf

lf

2

2

2

2

012

2 2

0

12

lf

lf

lf

l lf f

lf

F

Variational Method

Kd F

2 3

0 1 2 3( )v x c c x c x c x assume:

1 2

1 2

, (0) , ( ) ,

(0) , ( )

v v v l v

v v l

1v 2v( )v x

12

x

( )v xl

1v2v( )v x

1 0f

21

12fl21

12fl

x

( )v x

( ) :f x f const

1v2v( )v x

1

2fl

1

2fl

21

12fl21

12fl

2 0f

x

( )v x

( ) :f x f const

x

( )v x

( ) :f x f const

2

f l

1

2 2

1

3

2

2 2

2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

vl l

l l l lEI

l l vl

l l l l

F T

0, ( )

l

f dx F N

1

1

2

2

f

m

f

m

F

2

2

2

12

2

12

lf

lf

lf

lf

2

2 2

1

3

2 2 22

0

012 6 12 6

6 4 6 2 12

12 6 12 6 0 0

6 2 6 4

12

l l lf

l l l lEI

l ll

l l l l lf

2

2

20 2

l EI d vfv dx

dx

22/37



Seoul NationalUniv.

Element : Beam - Finite Element Methodequivalent nodal forces

Variational Method

1v2v( )v x

1 0f

21

12fl21

12fl

2 0f

x

( )v x

( ) :f x f const

2

2 2

1

3

2 2 22

0

012 6 12 6

6 4 6 2 12

12 6 12 6 0 0

6 2 6 4

12

l l lf

l l l lEI

l ll

l l l l lf

Kd F

givenfind

3 3

1 1 2 20, , 0,24 24

fl flv v

EI EI

,0 x l

displacement

given : x

find : ( )v x

3 2 2 3 3 2 2

1 23

1( ) ( 2 ) ( )

24

fv x x l x l xl x l x l

EI l 2 2 3( ) (2 )

24

fv x x l xl

EI

2

2

20 2

l EI d vfv dx

dx

23/37



Seoul NationalUniv.

Galerkin’s Residual Method

V

R dV

1

1

1 2 3 4

2

2

,

v

where N N N Nv

N d

Differential Equation4

4

( )0

d v xEI f

dx

( )u x Nd

test function

since it is approximated solution

4

4

( )d v xEA f

dx 0 R

residual

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

iN

0 ,( 1,2)i

V

R N dV i

1

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

basis function

24/37



Seoul NationalUniv.

Element : Beam - Galerkin’s Residual Method

the test functions are chosen to play the role of


Galerkin Method

iN

0 ,( 1,2)i

V

R N dV i weighting function

residual (test function used)N


4

4

( )0

d v xEI f

dx

4

40

( )0 , ( 1,2,3,4)

l

i

d v xEI f N dx i

dx

Beam - Galerkin’s Residual Method

integration by parts

3 3

3 30 00

0 ,( 1,2,3,4)

ll l

ii i

d v d v dNN EI EI dx f N dx i

dx dx dx

2 2 3 2

2 2 3 20 00

0 ,( 1,2,3,4)

ll l

i ii i

d N d v d v dN d vEI dx EI N f N dx i

dx dx dx dx dx

1


, ( )where v x Nd

3 2 3

1 3

1(2 3 )N x x l l

l

3 2 2 3

2 3

1( 2 )N x l x l xl

l

3 2

3 3

1( 2 3 )N x x l

l

3 2 2

4 3

1( )N x l x l

l

integration by parts again

2

20 00

0 ,( 1,2,3,4)

ll l

i ii i

d N dNEI dx EI N V m f N dx i

dx dx

B d

3 2

3 2( ), ( )

d v d vV x m x

dx dx

Recall,

0 00

0

lT

l lT T Td

EI dx EI m V f dxdx

NB B d N N

In matrix form,

25/37



Seoul NationalUniv.




Galerkin Method

iN

0 ,( 1,2)i

V




4

4

( )0

d v xEI f

dx

4

40

( )0 ,( 1,2)

l

i

d v xEI f N dx i

dx



1


, ( )where v x Nd

3 2 3

1 3

1(2 3 )N x x l l

l

3 2 2 3

2 3

1( 2 )N x l x l xl

l

3 2

3 3

1( 2 3 )N x x l

l

3 2 2

4 3

1( )N x l x l

l

2 2

3

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

l l

l l l lEI

l ll

l l l l

0

lTEI dx K B B2 2 2 3 2 2 2

3

( ) 16 6 3 4 6 6 3 2

d xx xl x l xl l x xl x l xl

dx l

N

3 2 3 3 2 2 3 3 2 3 2 2

3

1( ) 2 3 2 2 3x x x l l x l x l xl x x l x l x l

l N

(0) 1 0 0 0N

( )

, 0 0 0 1x l

d x

dx

N

0

( ), 0 1 0 0

x

d x

dx

N

, ( ) 0 0 1 0l N

0 00

( ) ( ) ( ) ( ) (0) (0) (0) (0)

lT T

l lT T T Td d d

EI m V f dx m l l V l l m V f dxdx dx dx

N N NN N N N N

R.H.S

0 00

0

lT

l lT T Td

EI dx EI m V f dxdx

NB B d N N

K

26/37



Seoul NationalUniv.




Galerkin Method

iN

0 ,( 1,2)i

V




4

4

( )0

d v xEI f

dx

4

40

( )0 ,( 1,2)

l

i

d v xEI f N dx i

dx



1


, ( )where v x Nd

3 2 3

1 3

1(2 3 )N x x l l

l

3 2 2 3

2 3

1( 2 )N x l x l xl

l

3 2

3 3

1( 2 3 )N x x l

l

3 2 2

4 3

1( )N x l x l

l

0 00

ll l

T dEI dx EI m V f dx

dx

NB B d N N

2 2

3

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

l l

l l l lEI

l ll

l l l l

0

lTEI dx K B B

R.H.S

0 00

2

2

( ) ( ) ( ) ( ) (0) (0) (0) (0)

20 0 0 1

0 0 1 0 12( ) ( ) (0) (0)

0 1 0 0

21 0 0 0

12

lT T

l lT T T Td d d

EI m V f dx m l l V l l m V f dxdx dx dx

lf

lf

m l V l m Vl

f

lf

N N N

N N N N N

2

2

(0)2

(0)12

( )2

( )12

lV f

lm f

lV l f

lm l f

Kd F2 2

3

2 2

12 6 12 6

6 4 6 2,

12 6 12 6

6 2 6 4

l l

l l l lEIwhere

l ll

l l l l

K

K

F (0) 1 0 0 0N

( )

, 0 0 0 1x l

d x

dx

N

0

( ), 0 1 0 0

x

d x

dx

N

, ( ) 0 0 1 0l N

2

2

(0)2

(0)12

,

( )2

( )12

lV f

lm f

lV l f

lm l f

F

27/37



Seoul NationalUniv.




Galerkin Method

iN

0 ,( 1,2)i

V




4

4

( )0

d v xEI f

dx

4

40

( )0 ,( 1,2)

l

i

d v xEI f N dx i

dx

1


, ( )where v x Nd


For simple support beam,

x

( )v x

( ) :f x f const

(0) , (0) 0, ( ) , ( ) 02 2

l lV f m V l f m l 2

1 2 3 4

2

0

12[ ]

0

12

T

lf

f f f f

lf

F

Kd F2 2

3

2 2

12 6 12 6

6 4 6 2,

12 6 12 6

6 2 6 4

l l

l l l lEIwhere

l ll

l l l l

K1 2 3 4, [ ]Tf f f fF

2

1 2

2

3 4

(0) , (0)2 12

( ) , ( )2 12

l lf V f f m f

l lf V l f f m l f

recall

1m 2m

28/37



Seoul NationalUniv.

Element : Beam - 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

29/37



Seoul NationalUniv.

Element : Beam - 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.

30/37



Seoul NationalUniv.

Element : Beam - 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

1

2 2

1 1

ext y s iy i i i

i iS

T v dS f v m

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

through displacement function

svsurface 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

yT

sv1S

nodal concentrated force moving through nodal displacements iviyf

l

1yf 2m

yT

x

,y v

2 yf

1m

31/37



Seoul NationalUniv.

Element : Beam - 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. 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.

iv

1

2 2

1 1

, ext y s iy i i i

i iS

T v dS f v m

32/37



Seoul NationalUniv.

l

1yf 2m

yT

x

,y v

2 yf

1m





in ext 1

2in x x

V

dV

Apply the following steps when using the principle of minimum potential

energy to derive the finite element equations.

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

sectional area is constantA

1

2 2

1 1

, ext y s iy i i i

i iS

T v dS f v m

dV dAdx

dS bdx

1

2 2

1 1

2 2

01 1

1

2

1

2

in ext

x x y s iy i i i

i iV S

l

x x y s iy i i i

i ix A

dV T v dS f v m

dAdx bT v dx f v m

The differential volume for the beam element

Width of beam, :b

The differential area over which the surface loading acts is

33/37



Seoul NationalUniv.





in ext 1

2in x x

V

dV

2 2

01 1

1

2

l

x x y s iy i i i

i ix A

dAdx bT v dx f v m



2

2x

du d vy

dx dx

1

2 2

1 1

, ext y s iy i i i

i iS

T v dS f v m

dvu y

dx

dv

dx

dv

dx

y

y Bd

x xE yE Bd

1

2 21

1 1 2 20 0

1 1 2

2

[ ]

y

l lT T T T T

y iy i i i y

i i y

f

mf dx f v m f dx v v

f

m

d N d N d F

T

02

lT TEI

dx d B Bd d F

0,

lT T

y s ybT v dx bT dx d N

l

1yf 2m

yT

x

,y v

2 yf

1m

y ybT f

34/37



Seoul NationalUniv.





in ext 1

2in x x

V

dV



T T T

02

lEIdx 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

0v

and

1

, 0

2 2

01 1

1

2

l

x x y s iy i i i

i ix A

dAdx bT v dx f v m

2

, 0v

2

, 0

1

2 2

1 1

, ext y s iy i i i

i iS

T v dS f v m

l

1yf 2m

yT

x

,y v

2 yf

1m

35/37



Seoul NationalUniv.





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

T T T

02

lEIdx d B Bd d F

1

2 2

1

1 1 2 23

2

2 2

2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

T T

vl l

l l l lEIv v u

l l vl

l l l l

d B Bd

00

lTEI dx B B d F

0d

1

1

02

2

,

y

l

y

y

f

mwhere f dx

f

m

F N

Kd F2 2

3

2 2

12 6 12 6

6 4 6 2,

12 6 12 6

6 2 6 4

l l

l l l lEIwhere

l ll

l l l l

K

1

1

02

2

,

y

l

y

y

f

mf dx

f

m

F Nx

( )v x

( ) :f x f const

36/37



Seoul NationalUniv.





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

Kd F2 2

3

2 2

12 6 12 6

6 4 6 2,

12 6 12 6

6 2 6 4

l l

l l l lEIwhere

l ll

l l l l

K

1

1

02

2

,

y

l

y

y

f

mf dx

f

m

F N

x

( )v x

( ) :f x f const

For simple support beam with uniform load f

,yf f 1

1

2

2

2

0

2

0

y

y

ff

m

ff

m

2 2

0

2

2

02 2

20

0 12 12,

0

2 2 2

0 12012

l

l lf f

f

l lf f

f dxf l l

f fl

fl

f

F N

37/37

Computer Aided Ship Design Part.3 Grillage Analysis of ...

Documents

Transcript of Computer Aided Ship Design Part.3 Grillage Analysis of ...