@ Computer Aided Ship Design

(1)

Naval Architecture & Oce an En gin ee ring

@ SDAL Advanced Ship Design Automation Lab.

http://asdal.snu.ac.kr Seoul

National Univ.

Computer Aided Ship Design

Part.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

(2)

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

@ SDAL

Seoul National

Application Bar

Beam

Shaft

Summary

Element Behavior

Tension

Bending

Torsion

Midship Cargo Hold

2 2

( ) ( ) 0 d u x

EA f x

dx  

4 4

( ) ( ) 0 EI w x f x

x

  



Structure

•Superposition of Stiffness Matrix

•Coordinates Transformation

Truss

Frame

Grillage

2 2

( ) x ( ) 0

GJ 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 , where x  0

Variational Method

2

0

( ) 0

2

l

EA du

f u dx

         dx         

2 2

0 2

( ) 0

2

l

EA d w

f w dx

     dx      

   

 



2

0

( ) 0

2

l

GJ d

f dx dx

                   

1 1

2 2

1 1

u f

EA

u f

l

    

      

  

     

0 1

( )

u x  a  a x

1 1

2 2

1 1 3

2 2

6 3 6 3

3 2 3

2

6 3 6 3

3 3 2

f u

l l

M

l l l l

EI

u

l l f

l

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 GJ M

l M



    

 

    

  

     

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

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Chapter 2. Element : Beam

(4)

@ SDAL

Seoul National

Element : Beam - Differential Eqn.

x  

ε i

① strain at in x-direction : y

x   , x  

σ i ε i , θ   k y ,  y j   E 

 ds

    d  ds d 1

ds



 

( y d ) d d y

ds y ds

    

 

  

     , : initial length

, :elongated length ds

y d  

ds

x y

neutral

surface y



dx

d 

ds 

σ x

* 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

(5)

Element : Beam - Differential Eqn.

x  

ε i

x ,

E where y

  

     

σ i i

( )

x x

d F  σ dA   i dA   dA i

③ 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 , θ   k y ,  y j   E 

 ds

    d  ds d 1

ds



 

x

E y

 

 σ  i   i d E y dA

 F    i

( y d ) d d y

ds y ds

    

 

  

     , : initial length

, :elongated length ds

y d  

neutral

surface y



dx

d 

ds ds 

x y

σ x

(6)

@ SDAL

Seoul National

Element : Beam - Differential Eqn.

 ds

    d  ds d 1

ds



 

① strain at in x-direction :

x  

ε i

x   , x  

σ i ε i

2

( ) ( y ) y

d d y E dA E dA

 

     

M y F j i k

y

, θ   k y ,  y j

( y d ) d d

ds y ds

    

       

  E 

② stress at in x-direction : y x y

 E

   

σ i i

③ force acting on in x-direction : dA d F   E  y dA i

2

A A

d E y dA

 M   M    k

⑤ assume  

dx dx dy

ds  ,   tan  

2 2

d d y

ds dx

 

2 2

d d dy d dy d y

ds ds dx dx dx dx

    

          



 A y dA

I 2

Define then, EI , EI

 M 

 

M k EI

 

M k

, d  : initial length, y d   :elongated length

2

, d y 2

M EI

 dx

2 2

EI d y

 dx

M k

EI d ds

 

M k

neutral

surface y



dx

d 

ds ds 

x y

σ x

6/37

(7)

Element : Beam - Differential Eqn.

1 d

ds



 

x  

ε i

x   , x  

σ i ε i

2

( ) ( y ) y

d d y E dA E dA

 

     

M y F j i k

y

, θ   k y ,  y j

( y d ) d d

ds y ds

    

       

  E 

 E

   

σ i i

2

A A

d E y dA

 M   M    k

⑤ assume  

dx dx dy

ds  ,   tan  

2 2

d d y

ds dx

 

, 2

I   A y dA

EI

  k

2 2

EI d y

 dx

M k

x y

( ) f x

neutral

surface y



dx

d 

ds ds

x y

σ x

EI d

EI ds



  

M k k

⑥ relationships between loads, shear forces, and bending moments

1 , 2 V , 1 , 2 M

V V dx M M dx

x x

 

   

                 

V j V j M k M k

•force equilibrium

1 2 ( ) 0

y    x 

 F V V f

 

1 1 ¹

 

1

( ) 0

V V V dx f x dx

x

V V V dx f x dx

    

              

      

  

j j j

j dV ( )

   f x

2

, d y 2

M EI

 dx

(8)

@ SDAL

Seoul National

Element : Beam - Differential Eqn.

1 d

ds



 

x  

ε i

x   , x  

σ i ε i

2

( ) ( y ) y

d d y E dA E dA

 

     

M y F j i k

y

, θ   k y ,  y j

( y d ) d d

ds y ds

    

       

  E 

 E

   

σ i i

2

A A

d E y dA

 M   M    k

⑤ assume  

dx dx dy

ds  ,   tan  

2 2

d d y

ds dx

 

, 2

I   A y dA

EI

  k

x y

( ) f x

neutral

surface y



dx

d 

ds ds

x y

σ x

1 , 2 V , 1 , 2 M

V V dx M M dx

x x

 

   

                 

V j V j M k M k

•force equilibrium dV ( )

dx   f x

dM ( ) dx V x

 

•moment equilibrium 1 2 2  

1 ( ) 0

z    d   2 d  x dx  

 M M M x V x f

  1  ( )  0

2

M V

M M dx dx V dx dx f x dx

x x

 

 

     

 k        k  i           j      i     j 

2 2

EI d y

 dx

M k

EI d

EI ds



  

M k k

2

, d y 2

M EI

 dx

8/37

(9)

Element : Beam - Differential Eqn.

1 d

ds



 

① strain at in x-direction :

x  

ε i

x   , x  

σ i ε i

2

( ) ( y ) y

d d y E dA E dA

 

     

M y F j i k

y

, θ   k y ,  y j

( y d ) d d

ds y ds

    

       

  E 

 E

   

σ i i

2

A A

d E y dA

 M   M    k

⑤ assume  

dx dx dy

ds  ,   tan  

2 2

d d y

ds dx

 

, 2

I   A y dA

EI

  k

x y

( ) f x

neutral

surface y



dx

d 

ds ds

x y

σ x

1 , 2 V , 1 , 2 M

V V dx M M dx

x x

 

   

                 

V j V j M k M k

•force equilibrium dV ( )

dx   f x dM ( )

dx  V x

•moment equilibrium

2 2

EI d y

 dx

M k

EI d

EI ds



  

M k k

2

, d y 2

M EI

 dx

2 2

d y M

dx  EI d y

³₃

1 dM 1 V x ( )

dx  EI  dx  EI 

⁴

4

1 1

d y dV ( )

dx  EI  dx   EI  f x

4

d y ( )

EI f x

  

(10)

@ SDAL

Seoul National

Element : Beam - Variational Method

Boundary condition

4

4 ( ) 0

EI d v f x

dx  

4

0 l d v 4 0

EI f v dx

dx 

 

 

 

 



multiply by and integrate  u

4 0 4

l d v

EI v f v dx

dx  

 

  

 



L.H.S:

   

3 3

3 0 3 0

0 l

l d v l

d v d v

EI v EI dx f v dx

dx dx dx

    

 

      

     

   

3

0 3 0

l d v d v l

EI dx f v dx

dx dx

 

 

    

 

 

 

2 2 2

2 0 2 2 0

0 l

l l

d v d v d v d v

EI EI dx f v dx

dx dx dx dx

  

 

     

   

integration by part

d d

v v

dx   dx

 

1 2

v v    2 v 

     

 

f v   fv

 

( ) ( )

b b

a h x dx a h x dx

 

   

operation



0

2 2

0

0 , 0

, 0, 0

x x l

v v

d v d v

EI EI

dx dx

 

 

2 2

0 0

l d v d v l

EI dx f v dx

dx dx

 

   

2 2 2 2

2 2 2

1 2

v v v

x x x

    

  

 

   

 

10/37

(11)

Element : Beam - Variational Method

Boundary Condition

4

4 ( ) 0

EI d v f x

dx  

4

0 l d v 4 0

EI f v dx

dx 

 

 

 

 



multiply by and integrate  u

4 0 4

l d v

EI v f v dx

dx  

 

  

 



L.H.S:

integration by part

d d

v v

dx   dx

 

1 2

v v    2 v 

     

 

f v   fv

 

2 2 2 2

2 2 2

1 2

v v v

x x x

    

  

 

   

 

( ) ( )

b b

h x dx h x dx

 

   

operation



0

2 2

0

0 , 0

, 0, 0

x x l

v v

d v d v

EI EI

dx dx

 

 

2 2

0 0

l d v d v l

EI dx f v dx

dx dx

 

   

 

2 2

0 0

1 2

l d v d v l

EI dx fv dx

dx dx

 

   

2  

0 2 2

l EI d v

fv dx

     dx     

 

 



(12)

@ SDAL

Seoul National

Element : Beam - Finite Element Method

Variational Method

x

( )

l v x

v 1 v x ( ) v 2

 1  2

2 3

0 1 2 3

( )

v x   c c x c x   c x

assume: , (0) v  v 1 , v l ( )  v 2

1 2

, dv (0) , dv ( )

dx   dx l  

discretization

1 element , 2 nodes finite element method

2  

2 0 2 2

l EI d v

fv dx

     dx     

   

 



12/37

(13)

Element : Beam - Finite Element Method

2 3

0 1 2 3 2

( )

v l   c c l  c l  c l  v (0) 0

v  c

0 1

c  v

   

2 2 1 2 1 2

3 1

2

c v v

l l  

    

Variational Method

v 1 v x ( ) v 2

 1  2 2 2 2  

0 2

l EI d v

fv dx

     dx     

   

 



2 3

0 1 2 3

( )

v x   c c x c x   c x

assume: , (0) v  v 1 , v l ( )  v 2

1 2

, dv (0) , dv ( )

dx   dx l  

(0) 1

dv c

dx 

2

1 2 3 2

( ) 2 3

dv l c c l c l

dx     

1 1

c   3 3  1 2  2  1 2 

2 1

c v v

l l  

   

(14)

@ SDAL

Seoul National

Element : Beam - Finite Element Method

0 1

c  v , c 2 3 2  v 1 v 2   1 2 1 2 

l l  

    

Variational Method

v 1 v x ( ) v 2

 1  2 2 2 2  

0 2

l EI d v

fv dx

     dx     

   

 



2 3

0 1 2 3

( )

v x   c c x c x   c x

assume: , (0) v  v 1 , v l ( )  v 2

1 2

, dv (0) , dv ( )

dx   dx l  

1 1

, c  

   

3 3 1 2 2 1 2

2 1

c v v

l l  

   

    2     3

1 1 2 1 2 1 2 3 1 2 2 1 2

3 1 2 1

( ) 2

v 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 2

3 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

(15)

 

1 1

1 2 3 4

2 2

( )

v v x N N N N

v



   

  

   

 

Variational Method

v 1 v x ( ) v 2

 1  2 2 2 2  

0 2

l EI d v

fv dx

     dx     

   

 



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

1 1 2 2

3 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 

(16)

@ SDAL

Seoul National

Element : Beam - Finite Element Method

 

1 2

1

1 2 3 4

2

2 2

( )

v d v x

B B B B dx v



   

  

   

   v x ( )  Nd ,

2 2

( ) d v x

dx  B d

 

1 1

1 2 3 4

2 2

( )

v v x N N N N

v



   

  

   

 

differentiation with respect to twice x

Variational Method

v 1 v x ( ) v 2

 1  2 2 2 2  

0 2

l EI d v

fv dx

     dx     

   

 



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

1 1 2 2

3 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 3 2

1 (6 4 )

B xl l

 l 

3 3

1 ( 12 6 )

B x l

 l   4 1 3 2

(6 2 )

B xl l

 l 

3 2 3

1 3

1 (2 3 )

N x x l l

 l   2 3 3 2 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

(17)

( )

v x  N d

2 2

( ) d v x

dx  B d

2 2

0 2 ( ) 0

2

l EI d v

fv dx

     dx      

   

 



 

0 0 ( ) 0

2

l T T l

EI dx f dx

      

  d B B d  Nd 

derivation

   

1 1

1 2 3 4 1 2 3 4

2 2

, ,

v

where N N N N B B B B

v



   

    

   

 

N B d

Variational Method v 1 v x ( ) v 2



1



²

2  

2 0 2 2

l EI d v

fv dx

     dx     

   

 



2 2

3 3 3 3

1 1 1 1

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

xl l l

x l l

xl l

  

 

  

 

   

 

  

 

(18)

@ SDAL

Seoul National

(derivation)

2 2

0 2 ( )

2

l EA d v

fv dx

     dx     

   

 



  T T

0 0 ( )

2

l l

T T

EI dx f dx

  

   

  d B Bd  d N 

 

T T T

0 0

1 ( )

2

l l

EI T dx f dx

      

   d    B B   d d     N    

T T

1

  2 

   

 d Kd d F 

   T    T

 d Kd  d F

    T 

 d Kd F 

 

0

l T

EI dx

 

K B B

  T T   T

1 1

2  2  

 d Kd  d K d  d F

1 1 2 2

[ ] T

d  v  v 

  T   T   T

1 1

2  2  

 d Kd  d Kd  d F    d T Kd  d K d T 

 T

K K

symmetry

 

0 0 ( )

2

l l

T T

EI dx f dx

  

   

  d B B d  N d 

 

^T ^T ^T

:

f Nd  Nd f  d N f Nd scalar

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 l

EI

l l

l

l l l l

  

  

 

     

  

 

T

0 l ( f dx )

  

F N

18/37

(19)

(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 l

l l l l l l l l

xl l x l xl l xl l xl l x l xl l xl l

l l l l l l l l

EI

x l

l

        



 

₃ ₃ ₃ ² ₃ ₃ ₃ ₃ ²

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 l

l l l l l l l

xl l x l xl l xl l xl l x l xl l xl l

l 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 l

EI

l l

l

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 l

l l l l

x l xl l x l xl l x l xl l x l xl l

l l l l

EI

x xl l

l 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 1

72 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 l

l l

x l xl l x l xl l x l xl l x l xl l

l l l l

 

 

         

 

          

 



 

0

l T

EI dx

 

K B B

(20)

@ SDAL

Seoul National

Element : Beam - Finite Element Method

    d T Kd F    0

-Chapter 2. Element : Beam

T

, F    0 l ( N f dx )

 Kd  F

( )

v x  N d

2 2

( ) d v x

dx  B d

2 2

0 2 ( ) 0

2

l EI d v

fv dx

     dx      

   

 



 

0 0 ( ) 0

2

l T T l

EI dx f dx

      

  d B B d  Nd 

derivation

   

1 1

1 2 3 4 1 2 3 4

2 2

, ,

v

where N N N N B B B B

v



   

    

   

 

N B d

Variational Method v 1 v x ( ) v 2



1



²

2  

2 0 2 2

l EI d v

fv 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 l

where EI

l l

l

l l l l

  

  

 

     

  

 

K

20/37

(21)

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

l l v

l l l l

EI

l l v

l

l l l l



  

 

      

  

      

    

   

F T

, F    0 l ( N f dx )

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

, (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

0 3 2

3

3 2 2

3

1 (2 3 )

1 ( 2 )

1 ( 2 3 )

1 ( )

l

x x l l l

x l x l xl

l f dx

x x l l

x l x l l

   

 

   

 

   

   

 

  

 



4

3 3

4 2

3 2 3

3 4

3

4 3

2

0

2 2

1 4 3 2

2

4 3

l

x x l xl

x x

l x l l

l x f

x l

x x

l l

   

 

   

     

   

 

   

     

   

    

     

 

1 1 2

2

f m

   

  

   

  F

v 1

v 2

( ) v x

1

2 fl 1

2 fl

 

2 2

0

2

2 l

EI d v

fv dx

     dx     

   

 



Variational Method

v 1 v

2

( ) v x



1



²

x

( )

l v x

x

( ) v x

( ) :

f x  f const

2

2 12

l f

  

 

  

 

  

  

 

 

1

2

12 fl 1

2

12 fl

(22)

@ SDAL

Seoul National

Element : Beam - Finite Element Method

constant external force per unit length

boundary condition equivalent nodal forces free body diagram

2 f l 

2

0

12 0

12 l f

l f

 

 

  

 

  

 

 

2

2 2 12 0

2 2

0 12

l f l f l f

l l

f f

l f

  

   

   

    

   

               

     

 

 

F

Variational Method

Kd  F

2 3

0 1 2 3

( )

v x   c c x c x   c x

assume:

1 2

, (0) , ( ) , (0) , ( )

v v v l v

v  v l 

 

   

v 1 v x ( ) v

₂



1



²

x

( )

l v x

v 1

v 2

( ) v x

1

0

f 

1

2

12 fl 1

2

12 fl

x

( ) v x

( ) :

f x  f const

v 1

v 2

( ) v x

1

2 fl 1

2

12 fl 1

2

12 fl

2

0

f 

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

l l v

l l l l

EI

l l v

l

l l l l



  

 

      

    

    

    

   

F

^T

, F   

0^l

( N f dx )

1 1 2

2

f m

   

  

   

  F

2

2 12

l f

  

 

  

 

  

  

 

 

2

2 2

1 3

2 2 2

2

0 0

12 6 12 6

6 4 6 2 12

12 6 12 6 0 0

6 2 6 4

l l l

l l l l f

EI

l l

l

l l l l l

f



 

 

  

      

      

      

        

    

   

 

 

 

2 2

0

2

2 l

EI d v

fv dx

     dx     

   

 



22/37

(23)

Variational Method

v 1

v 2

( ) v x

1

0

f 

1

2

12 fl 1

2

12 fl

2

0

f 

x

( ) v x

( ) :

f x  f const

2

2 2

1 3

2 2 2

2

0 0

12 6 12 6

6 4 6 2 12

12 6 12 6 0 0

6 2 6 4

12

l l l

l l l l f

EI

l l

l

l l l l l

f



 

 

  

 

  

 

      

      

        

    

   

 

 

Kd  F

given find

3 3

1

0,

1

,

2

0,

2

24 24

fl fl

v v

EI EI

 

    

, 0   x l

displacement given : x

find : v x ( )

3 2 2 3 3 2 2

1 3 2

( ) ( 2 ) 1 ( )

24

v x f x l x l xl x l x l

EI  l 

     ( ) (2 2 2 3 )

24

v x f x l xl

 EI 

 

2 2

0

2

2 l

EI d v

fv dx

     dx     

   

 



(24)

@ SDAL

Seoul National

Galerkin’s Residual Method

V

 R dV

 

1 1

1 2 3 4

2 2

,

v

where N N N N

v



   

     

   

N d

4 4

( ) 0

d v x

EI f

dx  

( )

u x  N d

test function since it is

approximated solution

4 4

( ) d v x

EA 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

0

V

R W dV 

 weighting function or

the basis functions are chosen to play the role of the weighting functions W

Galerkin Method

N i

0 , ( 1, 2)

i V

R N dV  i 



1

.)

0

( ) 0

ref   u v    uv  xv dx 

basis function

24/37

(25)

Element : Beam - Galerkin’s Residual Method

the test functions are chosen to play the role of the weighting functions W

Galerkin Method

N

i

0 , ( 1, 2)

i V

R N dV  i 

 weighting function

residual (test function used) N

Differential Equation

4 4

( ) 0

d v x

EI f

dx  

4 0 4

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

l

i

d v x

EI f N dx i

dx

 

  

 

 



Beam - Galerkin’s Residual Method

integration by parts

3 3

3 0 3 0

0

0 ,( 1, 2,3, 4)

l

l l

i

i i

d v d v dN

N EI EI dx f N dx i

dx dx dx

 

   

 

   

2 2 3 2

0 0

0

0 ,( 1, 2,3, 4)

l

l l

i i

d N d v d v dN d v

EI dx EI N f N dx i

dx dx dx dx dx

 

      

 

 

1

.)

0

( ) 0

ref   u v    uv  xv dx 

, ( )

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

0 2 0

0

0 , ( 1, 2,3, 4)

l

l l

i i

d N dN

EI dx EI N V m f N dx i

dx dx

 

        

 B d 

3 2

3 ( ), 2 ( )

d v d v

V x m x

dx  dx 

Recall,

0 0

0

T l

l l

T d T T

EI dx EI m V f dx

dx

 

     

 

 B B d N N  N

In matrix form,

(26)

@ SDAL

Seoul National

Element : Beam - Galerkin’s Residual Method

Galerkin Method

N

i

0 , ( 1, 2)

i V

R N dV  i 

4 4

( ) 0

d v x

EI f

dx  

4 0 4

( ) 0 ,( 1, 2)

l

i

d v x

EI f N dx i

dx

 

  

 

 



1

.)

0

( ) 0

ref   u v    uv  xv dx 

, ( )

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 l

EI

l l

l

l l l l

  

  

 

     

  

 

 

0

l T

EI dx

 

K B B

2 2 2 3 2 2 2

3

( ) 1

6 6 3 4 6 6 3 2

d x

x 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

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

l  

         

N

 

(0)  1 0 0 0 N

 

, ( ) 0 0 0 1

x l

d x dx  

  N

0

, ( ) 0 1 0 0

x

d x

dx   N

 

, ( ) N l  0 0 1 0

0 0

0

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

T l T

l l

T T T T

d d d

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

dx dx dx

 

      

 

 N N   N N N N N  N

R.H.S

0 0

0

T l

l l

T d T T

EI dx EI m V f dx

dx

 

     

 

 B B d N N  N

K

26/37

(27)

Element : Beam - Galerkin’s Residual Method

Galerkin Method

N

i

0 , ( 1, 2)

i V

R N dV  i 

4 4

( ) 0

d v x

EI f

dx  

4 0 4

( ) 0 ,( 1, 2)

l

i

d v x

EI f N dx i

dx

 

  

 

 



1

.)

0

( ) 0

ref   u v    uv  xv dx 

, ( )

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 0

0

l l l

T d

EI dx EI m V f dx

dx

 

      

 B B d N 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 l

EI

l l

l

l l l l

  

  

 

     

  

 

 

0

l T

EI dx

 

K B B

R.H.S

0 0

0

2

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

2

0 0 0 1

0 0 1 0 12

( ) ( ) (0) (0)

0 1 0 0

1 0 0 0 2

12

T l T

l l

T T T T

d d d

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

dx dx dx

l f

m l V l m V

l f

 

      

 

 

  

 

         

        

       

                                

  

 

N N N

N N N N N

2

(0) 2 (0) 12 ( ) 2 ( ) 12

V l f

m l f

V l l f

m l l f

  

  

    

  

   

    

  

   

    

 Kd  F

³ ² ²

12 6 12 6

6 4 6 2

, 12 6 12 6

l l

l l l l

where EI

l l

l

  

  

 

     

K

F  

(0)  1 0 0 0 N

 

, ( ) 0 0 0 1

x l

d x

dx

_



  N

0

, ( ) 0 1 0 0

x

d x dx

_

 N

 

, ( ) N l  0 0 1 0

2

(0) 2 (0) 12 ,

V l f

m l f

l

  

 

  

 

  

  

F

(28)

@ SDAL

Seoul National

Element : Beam - Galerkin’s Residual Method

Galerkin Method

N

i

0 , ( 1, 2)

i V

R N dV  i 

4 4

( ) 0

d v x

EI f

dx  

4 0 4

( ) 0 ,( 1, 2)

l

i

d v x

EI f N dx i

dx

 

  

 

 



1

.)

0

( ) 0

ref   u v    uv  xv dx 

, ( )

where v x  Nd

For simple support beam,

x

( ) v x

( ) :

f x  f const

(0) , (0) 0, ( ) , ( ) 0

2 2

l l

V  f m  V l   f m l  2

1 2 3 4

2

0

[ ] 12

0

12

T

l f f f f f

l f

 

 

  

 

   

 

 

F

 Kd  F

³ ² ²

2 2

12 6 12 6

6 4 6 2

, 12 6 12 6

6 2 6 4

l l

l l l l

where EI

l l

l

l l l l

  

  

 

     

  

 

K , F  [ f f 1 2 f 3 f 4 ] T

2

1 2

2

3 4

(0) , (0)

2 12

( ) , ( )

2 12

l l

f V f f m f

l l

f V l f f m l f

    

    

recall

m

1

m

²

28/37

(29)

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 Π in and the potential energy of the external forces Π ext

in ext

    

(30)

@ SDAL

Seoul National

Element : Beam - Potential Energy Approach

in ext

    

1

in in 2 x x

V V

d   dV

     

the strain energy for one-dimensional stress.

 x

 x

Linear-elastic

(Hooke’s law)material

x E x

  

0

x

in x

d       d dxdydz

d  x

0  x

0

x

x x

E d dxdydz

  

   1 2 E    x 2 d  x dxdydz

1

2  x dxdydz



To evaluate the strain energy for a bar,

we consider only the work done by the internal forces during deformation.

30/37

(31)

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 i

S

T v dS f v m 

 

       

b v

X

body forces typically from the self-weight of the bar (in units of force per unit volume) moving through displacement function

v s

surface 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 T y

v s

S

1

nodal concentrated force moving through nodal displacements f iy v i

l

f 1y m 2

T y

x ,

y v

2 y

f

m 1

(32)

@ SDAL

Seoul National

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 Π

_in

and the potential energy of the external forces Π

_ext

in ext

      

ⁱⁿ

1 2 

_V

 

^x ^x

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.

v i

1

2 2

1 1

,

_ext _{y s} _{iy i} _{i i}

i i

S

T v dS f v m 

 

       

32/37

(33)

l

f

1y

m

²

T

y

x ,

y v

2y

f m

1

the principle of minimum potential energy

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

_in

and the potential energy of the external forces Π

_ext

in ext

      

ⁱⁿ

1 2 

_V

 

^x ^x

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 constant A

1

2 2

1 1

,

_ext _{y s} _{iy i} _{i i}

i i

S

T v dS f v m 

 

       

dV  dAdx dS  bdx

1

2 2

1 1

2 2

0 1 1

1 2 1 2

in ext

x x y s iy i i i

i i

V S

l

x x y s iy i i i

i i

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