Innovative Ship Design - Elasticity
@
SDALAdvanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul
National Univ.
N a v a l A rc hi te ct u re & Oc e a n En g in e e ri n g
@ SDALAdvanced Ship Design Automation Lab.
http://asdal.snu.ac.kr Seoul
National Univ.
June 2009
Prof. Kyu-Yeul Lee
Department of Naval Architecture and Ocean Engineering, Seoul National University of College of Engineering
[2009] [15]
Innovative ship design
- Variational Mehod and Approximation-
서울대학교 조선해양공학과 학부4학년 “창의적 선박설계” 강의 교재
weak form :
‘weighted average’*
Contents
Approximation
Governing Equation
Differential Approach Energy Method
Stress
Strain Displacement
Force Equilibrium
Generalized Hooke’ Law Compatibility
Analytic Solution
Elasticity
Strain Energy , 0
=m =
∑
F a a in equilibriumVirtual Displacement Virtual Work
FEM
Rayleigh-Ritz Method Galerkin Method Deflection Curve of the Beam
Scantling
2
Wang,C.T.,
Applied Elasticity , McGRAW-HILL, 1953 응용탄성학, 이원 역, 숭실대학교 출판부, 1998 Chou,P.C.,
Elasticity (Tensor, Dyadic, and Engineering Approached), D. Van Nostrand, 1967
Gere,J.M.,
Mechanics of Materials, Sixth Edition, Thomson, 2006
Hildebrand,F.B.,
”Methods of Applied Mathematics”, 2ndedition, Dover, 1965 Becker,E.B.,
“Finite Elements, An Introduction”, Vol.1, Prentice-Hall, 1981 Fletcher,C.A.J.,
“Computational Galerkin Methods”, Springer, 1984
1 2
1
Innovative Ship Design - Elasticity
@
SDALAdvanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul
National Univ.
Summary
( ) e 2 0
G G u X
λ+ ∂x+ ∇ + =
∂
( ) e 2 0
G G v Y
λ+ ∂y+ ∇ + =
∂
( ) e 2 0
G G w Z
λ+ ∂z + ∇ + =
∂
(1 )(1 2 ) (1 ) (1 )(1 2 ) (1 ) (1 )(1 2 ) (1 )
x x
y y
z z
E E
e
E E
e
E E
e
σ ν ε
ν ν ν
σ ν ε
ν ν ν
σ ν ε
ν ν ν
= +
+ − +
= +
+ − +
= +
+ − +
, e = ε
x+ ε
y+ ε
z, 2( 1)
, 2( 1)
, 2( 1)
xy xy
yz yz
zx zx
E E E
τ γ
ν
τ γ
ν
τ γ
ν
= +
= +
= +
6 Relations btw. 6 Strain and 6 Stress 6 Relations btw. Strain and Displacement
, , ,
, ,
x y z
xy yz zx
u v w
x y z
u v v w w u
y x z y x z
ε ε ε
γ γ γ
∂ ∂ ∂
= = =
∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂
= + = + = +
∂ ∂ ∂ ∂ ∂ ∂
6 Equations of force equilibrium 0
0
0
x yx zx
x
xy y zy
y
xz yz z
z
F X
x y z
F Y
x y z
F Z
x y z
σ τ τ
τ σ τ
τ τ σ
∂ ∂ ∂
= + + + =
∂ ∂ ∂
∂ ∂ ∂
= + + + =
∂ ∂ ∂
∂ ∂ ∂
= + + + =
∂ ∂ ∂
∑
∑
∑
0 0 0
x yz zy
y xz zx
z xy yx
M M M
τ τ τ τ τ τ
= − =
= − =
= − =
∑ ∑
∑
, , , , , , , ,
x yx zx xy y zy xz yz z
σ τ τ τ σ τ τ τ σ
, , u v w
, , , , ,
x y z xy yz zx
ε ε ε γ γ γ
6 Strain 9 Stress 3 Displacement
18 Variables
, , u v w
Given : Body force
X Y Z , ,
Find : Displacement
2 1 3
18 Equations
3 Variables 3 Equations
If we are interested in finding the displacement components in a body, we may reduce the system of equations to three equations with three unknown displacement components.
Variables and Equations
, : Lame Elastic constant µ λ
: Shear Moldulus G
: Young's Modulus E
u v w
e x y z
∂ ∂ ∂
= + +
∂ ∂ ∂
, , : bodyforce in x,y, and z direction repectively X Y Z
: Poisson's Ratio ν
2 2 2
2
2 2 2
x z y
∂ ∂ ∂
∇ = + +
∂ ∂ ∂
x y z
σ σ σ
Θ = + +
3/183
Summary
Given : Body force
(1 )(1 2 ) (1 ) (1 )(1 2 ) (1 ) (1 )(1 2 ) (1 )
x x
y y
z z
E E
e
E E
e
E E
e
σ ν ε
ν ν ν
σ ν ε
ν ν ν
σ ν ε
ν ν ν
= +
+ − +
= +
+ − +
= +
+ − +
ε ε ε
= + +
, 2( 1)
, 2( 1)
, 2( 1)
xy xy
yz yz
zx zx
E E E
τ γ
ν
τ γ
ν
τ γ
ν
= +
= +
= +
6 Relations btw. 6 Strain and 6 Stress 6 Equations of force equilibrium
0 0
0
x yx zx
x
xy y zy
y
xz yz z
z
F X
x y z
F Y
x y z
F Z
x y z
σ τ τ
τ σ τ
τ τ σ
∂ ∂ ∂
= + + + =
∂ ∂ ∂
∂ ∂ ∂
= + + + =
∂ ∂ ∂
∂ ∂ ∂
= + + + =
∂ ∂ ∂
∑
∑
∑
0 0 0
x yz zy
y xz zx
z xy yx
M M M
τ τ τ τ τ τ
= − =
= − =
= − =
∑ ∑
, , ∑
X Y Z
2
1 3
15 Equations
2 2
2 2
2 2
1 0
1
1 0
1
1 0
1
xy
yz
zx
Y X
x y x y
Z Y
y z y z
X Z
z x z x
τ ν
τ ν
τ ν
∂ ∂ ∂ Θ
+ + ∇ + =
∂ ∂ + ∂ ∂
∂ ∂ ∂ Θ
+ + ∇ + =
∂ ∂ + ∂ ∂
∂ ∂ ∂ Θ
+ + ∇ + =
∂ ∂ + ∂ ∂
2 2
2
2 2
2
2 2
2
2 1 0
1 1
2 1 0
1 1
2 1 0
1 1
x
y
z
X Y Z X
x y z x x
X Y Z Y
x y z y y
X Y Z Z
x y z z z
ν σ
ν ν
ν σ
ν ν
ν σ
ν ν
∂ ∂ ∂ ∂ ∂ Θ
+ + + + ∇ + =
− ∂ ∂ ∂ ∂ + ∂
∂ +∂ +∂ + ∂ + ∇ + ∂ Θ=
− ∂ ∂ ∂ ∂ + ∂
∂ +∂ +∂ + ∂ + ∇ + ∂ Θ=
− ∂ ∂ ∂ ∂ + ∂
, , , , ,
x y z xy yz zx
σ σ σ τ τ τ
Find : Stress
6 Variables 6 Equations
, , , , , , , ,
x yx zx xy y zy xz yz z
σ τ τ τ σ τ τ τ σ
, , u v w
, , , , ,
x y z xy yz zx
ε ε ε γ γ γ
6 Strain 9 Stress 3 Displacement
15 Variables
6 Relations btw. Strain and Displacement
, , ,
, ,
x y z
xy yz zx
u v w
x y z
u v v w w u
y x z y x z
ε ε ε
γ γ γ
∂ ∂ ∂
= = =
∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂
= + = + = +
∂ ∂ ∂ ∂ ∂ ∂
2
2
2
2
2
2
yz xy
x zx
y yz zx xy
yz zx xy
z
y z x x y z
or z x y x y z
x y z x y z
γ γ
ε γ
ε γ γ γ
γ γ γ
ε
∂ = ∂ −∂ +∂ +∂
∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂
= − +
∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂
∂ ∂ =∂ ∂ + ∂ − ∂
2 2
2
2 2
2 2 2
2 2
2 2
2
2 2
y xy
x
y z yz
x zx
z
y x x y
z y y z
x z z x
ε γ
ε
ε ε γ
ε γ
ε
∂ +∂ =∂
∂ ∂ ∂ ∂
∂ ∂ ∂
+ =
∂ ∂ ∂ ∂
∂ +∂ =∂
∂ ∂ ∂ ∂
Compatibility equations 3 independent Equations
18 Variables
18 Equations
If we are interested in finding only the stress components in a body, we may reduce the system of equations to six equations with six unknown stress components
, : Lame Elastic constant µ λ
: Shear Moldulus G
u v w
e x y z
∂ ∂ ∂
= + +
∂ ∂ ∂
, , : bodyforce in x,y, and z direction repectively X Y Z
: Poisson's Ratio ν
2 2 2
2
2 2 2
x z y
∂ ∂ ∂
∇ = + +
∂ ∂ ∂
x y z
σ σ σ
Θ = + +
Innovative Ship Design - Elasticity
@
SDALAdvanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul
National Univ.
Differential Equation (ODE/PDE)
Classification
Integral Equations
Variational formulation
Rayleigh-Ritz Approximate Method
Galerkin Collocation
Least Square
Approximate Method
Volterra
Weak Form2) Approximate Method4)
Galerkin Collocation Least Square
FEM
( ) ( ) ( ) x ( , ) ( )
x y x F x aK x y d
α = +λ∫ ξ ξ ξ
( ) ( ) ( ) b ( , ) ( )
x y x F x aK x y d
α = +λ∫ ξ ξ ξ
Fredholm Leibnitz formula1)
( ) ( )
( ) ( )
( , )
( , ) [ , ( )] [ , ( )]
B x B x
A x A x
d d F x dB dA
F x d d F x B x F x A x
dx dx x dx dx
ξ ξ= ∂ ξ ξ+ −
∫ ∫ ∂
1) Jerry, A.j., Introduction to Integral Equations with Applications, Marcel Dekker Inc., 1985, p19~25
2) ‘variational statement of the problem’ -Becker, E.B., et al, Finite Elements An Introduction, Volume 1, Prentice-Hall, 1981, p4
3) Becker, E.B., et al, Finite Elements An Introduction, Volume 1, Prentice-Hall, 1981, p2 . See also Betounes, Partial Differential Equations for Computational Science, Springer, 1988, p408 “…the weak solution is actually a strong (or classical) solution…”
4) some books refer as ‘Method of Weighted Residue’ from the Finite Element Equation point of view and they have different type depending on how to choose the weight functions. See also Fletcher,C.A.J., “Computational Galerkin Methods”, Springer, 1984
5) Jerry, A.j., Introduction to Integral Equations with Applications, Marcel Dekker Inc., 1985, p1 “Problems of a ‘hereditary’ nature fall under the first category, since the state of the system u(t) at any time t depends by the definition on all the previous states u(t-τ) at the previous time t-τ ,which means that we must sum over them, hence involve them under the integral sign in an integral equation.
whenever a smooth ‘classical(strong)’
solution to a (D.E.) problem exists, it is also the solution of the weak
problem3) 1
0( ) 0
(0 )0, (1) 0 u u x v dx
u u
− + −′′ =
= =
∫
10(−u v′ ′+uv−xv dx) =0
∫
integration by part and demand the test functions
vanish at the endpoints
2 0
d dy
T y p
dx dx + ρω + =
Ex.)
2 2 2
0
1 0
2 2
l T dy
y py dx
δ∫ ρω + − dx =
, 0 1, (0 )0, (1) 0
u u x x
u u
− + =′′ < <
= =
Ex.) Work and Energy Principle
1
( ) ( )
n
k k i i
k
c s x F x
=
∑
=( ) ( ) 2
1
min
b n
k k
a k
c s x F x dx
=
−
∑
∫
1
( ) ( ) ( ) ( )
n b b
k a i k a i
k
c ψ x s x dx ψ x F x dx
=
∑ ∫ =∫
, ( )ψ x =
∑
nk=1ak iφ( )x1
( ) ( )
n k k k
c s x F x
=
∑
≈ ,s xk( )=φk( )x −λ∫abK x( , ) ( )ξ yξ ξdassume:
0 1 1
( ) ( ) ( ) n n( ) y x ≈φ x +cφ x ++cφ x
problem of a “hereditary’ nature5) multiply and integration δy
integration by part and B/C
2 0
l d dy
T y p y dx
dx dx ρω δ
+ +
∫
multiply and integration v
( ) nj 1 j j( ) u x ≈
∑
= cφ x( ) n1 i i( ) v x ≈
∑
i= aφ x- Variation and integration - Integration and variation
( ) ( ) ( ) ( )
{
01}
01 ( )1 1 1
N N N
i i i j i j i i
i j i
a c φ x φ x φ x φ x dx a xφ x dx
= = =
′ ′ + =
∑ ∑ ∫ ∑ ∫
1 n
j ij i
i= c k =F
∑
kij
Fi
1 2 3 4 x
1( )x φ
( )
2 x φ
3( )x φ
shape function
what is the relationship between ‘week form’ and
‘Variational formulation’?
[ ]
1 1 1
0(−u v dx′′ ) = −u v′ 0+ 0(u v dx′ ′)
∫ ∫
5/183
Classification
what is the relationship between ‘week form’ and ‘Variational formulation’?
Differential Equation (ODE/PDE)
Integral Equations
Variational formulation
Rayleigh-Ritz Approximate Method
Galerkin Collocation
Least Square Approximate Method
Volterra
Weak Form2) Approximate Method4)
Galerkin Collocation Least Square
FEM
( ) ( ) ( ) x( , ) ( ) a
x y x F x K x y d
α = +λ∫ ξ ξ ξ
( ) ( ) ( ) b ( , ) ( ) a
x y x F x K x y d
α = +λ∫ ξ ξ ξ
Fredholm Leibnitz formula1)
( ) ( )
( ) ( )
( , )
( , ) [ , ( )] [ , ( )]
B x B x
A x A x
d d F x dB dA
F x d d F x B x F x A x
dx dx x dx dx
ξ ξ= ∂∂ξ ξ+ −
∫ ∫
whenever a smooth ‘classical(strong)’
solution to a (D.E.) problem exists, it is also the solution of the weak
problem3) 1
0( ) 0
(0 )0, (1) 0 u u x v dx
u u
− + −′′ =
= =
∫ 1
0(−u v′ ′+uv−xv dx) =0
∫
alternative form (integration by part and B/C)
2 0
d dy
T y p
dx dx + ρω + =
Ex.)
2 22 0
1 0
2 2
l T dy
y py dx
δ∫ρω + − dx = , 0 1,
(0 )0, (1) 0
u u x x
u u
− + =′′ < <
= =
Ex.) Work and Energy Principle
1
( ) ( )
n
k k i i
k
c s x F x
=
∑ =
( ) ( )2
1 min
bn kk a
k
c sx F x dx
=
−
∑
∫
1
( )( ) ( ) ( )
n b b
kai k ai
k
c ψx s x dx ψx F x dx
=
∑ ∫ =∫
, ( ) n1k i( )
x ka x
ψ =∑= φ
1
( ) ( )
n k k k
c s x F x
=
∑ ≈ ,s xk( )=φk( )x−λ∫abK x( , ) ( )ξyξ ξd assume:
0 1 1
( ) ( ) ( ) n n( )
y x≈φ x+cφx++cφ x
problem of a “hereditary’ nature5) multiply and integration δy
integration by part and B/C 2
0
l d dy
T y p y dx
dx dx ρω δ
+ +
∫
multiply and integration v
( ) nj1jj( ) u x≈∑=cφ x ( ) n1i i( ) v x≈∑i=aφx
- Variation and integration - Integration and variation
1 n
j ij i
i=c k =F
∑
1 2 34
shape function
d dy 2
T y p
dx dx + ρω = −
Ex: Rotating String)
2 2 2
0
1 0
2 2
l T dy
y py dx
δ
∫
ρω + − dx =differential equation
2 2 2
0 0
1 0
2 2
l l
T dy y y T dy
py dx d
x
ρ dx δ
δ ω + − + =
∫
2
0l d dy 0
T y p dx
dx d y
x ρω δ
+ + =
∫
multiply and integration δy
integration by part and B/C
Variational formulation:
physical meaning : “Minimize the difference of kinetic energy and
potential energy”
2
0 0
l d dy l
T y dx p dx
dx dx ρω ν ν
+ = −
∫ ∫
multiply and ‘week form’
ν
Weighted Residual
if is has a meaning of minimizing the weighted residual, we may rewrite is as
virtual displacement (it has a physical meaning)
self-adjointform Natural B/C
2
0l d dy 0
T y p dx
dx dx ρω ν
+ + =
∫
Ritz method ( ) n 1 j j( ) y x ≈
∑
j= cφ x( ) nj 1 j j( ) y x ≈
∑
= cφ x( ) n1 i i( ) v x ≈
∑
i= aφ x2
0l d dy 0
T y p dx
dx dx
δ
∫
+ ρω + ν =in case of using Galerkin method, weight function can be regarded as a kind of y since they have same basis functions
2 0
l d dy
T y p ydx
δ
∫
+ ρω + =a similarity
Innovative Ship Design - Elasticity
@
SDALAdvanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul
National Univ.
Classification
1) Becker E.B., Finite Elements An Introduction, Volume1, Prentice-Hall, 1981
what is the relationship between ‘week form’ and ‘Variational formulation’?
Our reference to certain weak forms of boundary-value problems as
“variational” statements arises from the fact that, whenever the
operators involved possess a certain symmetry, a weak form of the problem can be obtained which is precisely that arising in standard problems in the calculus of variations.
7/183
Classification
what is the relationship between ‘week form’ and ‘Variational formulation’?
Remark 14.11)In the case of positive definite operators, the Galerkin method brings nothing new in comparison with the Ritz method ; the two methods lead to the solution of identical systems of linear equations and to identical sequences of approximate solution. However, the possibility of the application of the Galerkin method is much broader that that of the Ritz method.
(Aun − f,ϕk)=0, k=1,...,n
The Galerkin method, which is characterized by the condition does not impose beforehand any essential restrictive conditions on operator
It is in no way necessary that the operator A be positive definite, it need not even be symmetric, above all it need not ne linear.
Formally speaking, the Galerkin method can thus applied even in the case of very general operators
A
Remark 14.2.Although both the Ritz and the Galerkin methods lead to the same results in the case of linear positive operators, the basic ideas of these methods are entirely different.
ex: deflection of beam
[
EIu′′ =]
′′ q x( ) with the B/C u(0) =u′(0) =0, ( )u l =u l′( )=0,2
0 0
1 ( )
2
l l
EI u′′ dx− q u dx
∫ ∫
or minimize the function of energy
Definition 8.152) An operator is called positive in its domain if it is symmetric and if for all , the relations A DA u∈DA (Au u, )≥0 and (Au u, )=0 ⇒ =u 0 hold.
( , ) b 2
Au u =
∫
a Au dx′If, moreover, there exists a constant such that for all the relation holds, then the operator is called positive definite in
0
C > u∈DA (Au u, )≥C u2 2
A DA
DA
in
Innovative Ship Design - Elasticity
@
SDALAdvanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul
National Univ.
Classification
1) Rectorys, K. Variational Methods in Mathematics, Science and Engineering, Second edition, D.Reidel Publishing, 1980, p163
what is the relationship between ‘week form’ and ‘Variational formulation’?
Remark 14.2.Although both the Ritz and the Galerkin methods lead to the same results in the case of linear positive operators, the basic ideas of these methods are entirely different.
ex: deflection of beam
[
EIu′′ =]
′′ q x( ) with the B/C u(0) =u′(0) =0, ( )u l =u l′( )=0, 20 0
1 ( )
2
l l
EI u′′ dx− q u dx
∫ ∫
or minimize the function of energy
1 ,
n
n k i i
u =
∑
= aϕapproximate solution where satisfy the B/Cϕi
Galerkin method
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
1 0 1 1 2 0 2 1 0 1 0 1
1 0 1 2 2 0 2 2 0 2 0 2
1 0 1 2 0 2 0 0
l l l l
n n
l l l l
n n
l l l l
n n n n n n
a EI dx a EI dx a EI dx q dx
a EI dx a EI dx a EI dx q dx
a EI dx a EI dx a EI dx q dx
ϕ ϕ ϕ ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ ϕ ϕ ϕ
′′ ′′ ′′
′′ + ′′ + + ′′ =
′′ ′′ ′′
′′ + ′′ + + ′′ =
′′ ′′ ′′
′′ + ′′ + + ′′ =
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
: multiply and integration
ϕ
Ritz method
( )
( )
( )
2
1 0 1 2 0 1 2 0 1 0 1
2
1 1 2 2 2 2 2
0 0 0 0
2
1 0 1 2 0 2 0 0
l l l l
n n
l l l l
n n
l l l l
n n n n n
a EI dx a EI dx a EI dx q dx
a EI dx a EI dx a EI dx q dx
a EI dx a EI dx a EI dx q dx
ϕ ϕ ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ ϕ ϕ
′′ + ′′ ′′ + + ′′ ′′ =
′′ ′′ + ′′ + + ′′ ′′ =
′′ ′′ + ′′ ′′ + + ′′ =
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
∫ ∫ ∫ ∫
different?
The Galerkin method starting with the differential equation of the problem and the Ritz method with the respective functional
( ) ( ) ( ) ( ) ( )
( )
0 0 0
0
0
l l l l
i k i k i k i k i k
l
i k
EI dx EI EI dx EI EI dx
EI dx
ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ
ϕ ϕ
′′ ′ ′
′′ = ′′ − ′′ ′ = − ′′ ′+ ′′ ′′
′′ ′′
=
∫ ∫ ∫
∫
identical9/183