FINITE ELEMENTS I - Instituto Tecnológico de Aeronáuticaarfaria/AE245_08.pdf · Instituto Tecnológico de Aeronáutica AE-245 3 • Euler-Bernoulli beam model and Kirchhoff plate

Instituto Tecnológico de Aeronáutica

AE-245 1

FINITE ELEMENTS I

Class notes


AE-245 2

8. Beams and Plates


AE-245 3

• Euler-Bernoulli beam model and Kirchhoffplate model require C1 continuity

• For plate elements C1 continuity elements are difficult to implement

• Propose elements that require only C0

continuity and accommodate transverse shear

BEAMS AND PLATESIntroduction


AE-245 4

Beams


AE-245 5

Timoshenko Beams

Assumptions

1. Domain: Ω = Ω1 ∪ Ω2 ∪ … ∪ Ωn

Ωi = (x,y,z)∈ℜ3| x∈[0,hi], (y,z)∈Ai⊂ℜ2

1

2

34

5

671

2

3

4

5

6

7

x

y

z

θxθy

θz

h

A


AE-245 6

2. σyy = σzz = τyz = 0

Assumption used in the constitutive equation to eliminate εyy, εzz and γyz.

Timoshenko Beams


AE-245 7

3. Kinematic relations

)()(),,(

)()(),,(

)()()(),,(

xyxwzyxw

xzxvzyxv

xzxyxuzyxu

x

x

yz

θθ

θθ

+=−=

+−=

Warping is not included, that is, plane sections remain plane after deformation.

Timoshenko Beams


AE-245 8

Constitutive equation (isotropic)

−−−−−−

=

zz

yy

xx

zz

yy

xx

Eσσσ

νννννν

εεε

1

1

11

=

yz

xz

xy

yz

xz

xy

Gτττ

γγγ

100

010

0011

0=== yzzzyy τσσ

xzxz

xyxy

xxxx

G

G

E

γτγτεσ

=

==

Timoshenko Beams


AE-245 9

Strain -displacement relations

yxxxxzxz

zxxxxyxy

xyxzxxxx

ywwu

zvvu

zyuu

θθγθθγ

θθε

++=+=

−−=+=

+−==

,,,,

,,,,

,,,,

Timoshenko Beams


AE-245 10

∫

∫

∫

−=

=

−=

A

xxzz

A

xxyy

A

xyxzxx

dAyM

dAzM

dAzyM

σ

σ

ττ )(

Moment Force

∫

∫

∫

=

=

=

A

xzzz

A

xyyy

A

xxxx

dAQ

dAQ

dAQ

τ

τ

σ

Timoshenko Beams


AE-245 11

Prescribed displacements and rotations

zzyyxx

WwVvUu

Θ=Θ=Θ====

)0(,)0(,)0(

)0(,)0(,)0(

θθθ

Prescribed transverse force per unit length:

zyx qqq ,,

Timoshenko Beams


AE-245 12

zzyyxx QQQ ,,Prescribed boundary forces:

Qyy

y

x

z

Qzz

Qxx

Timoshenko Beams


AE-245 13

zzyyxx MMM ,,Prescribed boundary moment:

Myy

y

x

z

Mzz

Mxx

Timoshenko Beams


AE-245 14

Area properties

0

0

0

=

=

=

∫

∫

∫

A

A

A

yzdA

zdA

ydA

zzyy

A

xx

A

zz

A

yy

IIdAzyJ

dAyI

dAzI

+=+=

=

=

∫

∫

∫

)( 22

2

2

Timoshenko Beams


AE-245 15

Equilibrium equation: forces

y

x

z

Qxx + dQxx

Timoshenko Beams

Qxx

Qzz + dQzzQyy + dQyy

Qzz

Qyy

dx

qzz

qyy

qxx

0

0

0

,

,

,

=+

=+=+

zxzz

yxyy

xxxx

qQ

qQ

qQ


AE-245 16

Equilibrium equation: moment about x

y

x

z

Mxx + dMxx

Timoshenko Beams

Mxx

dx 0, =xxxM


AE-245 17

Equilibrium equation: moment about y

y

x

z

Timoshenko Beams

Qzz + dQzzMyy + dMyy

Qzz

Myy

dx

qzz0, =− zzxyy QM


AE-245 18

Equilibrium equation: moment about z

y

x

z

Timoshenko Beams

Mzz + dMzzQyy + dQyy

Mzz

Qyy

dx

qyy

0, =+ yyxzz QM


AE-245 19

Strong form

Timoshenko Beams

0

0

0

,

,

,

=+

=+=+

zxzz

yxyy

xxxx

qQ

qQ

qQ

0

0

0

,

,

,

=+

=−=

yyxzz

zzxyy

xxx

QM

QM

M

zzzz

yyyy

xxxx

zzzz

yyyy

xxxx

MhM

MhM

MhM

QhQ

QhQ

QhQ

=

=

=

=

=

=

)(

)(

)(

)(

)(

)(

zz

yy

xx

Ww

Vv

Uu

Θ=

Θ=Θ=

===

)0(

)0(

)0(

)0(

)0(

)0(

θθθ

boundary conditions


AE-245 20

Weak form

0)()(

)()()(

0

,

0

,

0

,

0

,

0

,

0

,

=++−+

++++++

∫∫∫

∫∫∫h

zyyxzz

h

yzzxyy

h

xxxx

h

zxzz

h

yxyy

h

xxxx

dxQMdxQMdxM

wdxqQvdxqQudxqQ

δθδθδθ

δδδ

Timoshenko Beams

δu, δv, δw, δθx, δθy, δθz satisfy geometric homogeneous boundary conditions


AE-245 21

Integration by parts

0)()()(

)()()()(

)(

)(

0

0

,,,

0

,,,

=−−

−−−−++−

+++−+

+++

∫

∫

∫

hMhMhM

hwQhvQhuQdxwqvquq

dxQwQQMvQ

dxMMuQ

zzzyyyxxx

zzyyxx

h

zyx

h

yzzxzzzyyxxxxxyy

h

xyyyxzzzxxx

δθδθδθ

δδδδδδ

δθδδθδθδ

δθδθδ

Timoshenko Beams


AE-245 22

Weak form: given prescribed displacements, rotations and forces find u, v, w, θx, θy, θz ∈ Ssuch that for all δu, δv, δw, δθx, δθy, δθz ∈ V

0)()()(

)()()()(

)(

)(

0

0

,,,

0

,,,

=−−

−−−−++−

−++−+

+++

∫

∫

∫

hMhMhM

hwQhvQhuQdxwqvquq

dxQwQQMvQ

dxMMuQ

zzzyyyxxx

zzyyxx

h

zyx

h

yzzxzzzyyxxxxxyy

h

xyyyxzzzxxx

δθδθδθ

δδδδδδ

δθδδθδθδ

δθδθδ

Timoshenko Beams


AE-245 23

Introduction of constitutive and area relations:

0)()()(

)()()()(

][

)]()()()[(

0

0

,,,,,,,,

0

,,,,

=−−

−−−−++−

++++

++++−−

∫

∫

∫

hMhMhM

hwQhvQhuQdxwqvquq

dxGJEIEIEAuu

dxwGAwvGAv

zzzyyyxxx

zzyyxx

h

zyx

h

xxxxxxxzzzxzxyyyxyxx

h

yxyxzxzx

δθδθδθ

δδδδδδ

θδθθδθθδθδ

θδθδθδθδ

Timoshenko Beams


AE-245 24

Introduce shape functions

∑=

=m

iii

h uNu1

∑=

=m

ixii

hx N

1

θθ

∑=

=m

iii

h wNw1

∑=

=m

iii

h vNv1

∑=

=m

iyii

hy N

1

θθ ∑=

=m

izii

hz N

1

θθ

Use same shape functions for δu, δv, δw, δθx, δθy, δθz

Timoshenko Beams


AE-245 25

Define vectors and matrices

Tzmymxmmmmzyx wvuwvud θθθθθθ L111111 =

Tzmymxmmmmzyx wvuwvud δθδθδθδδδδθδθδθδδδδ L111111 =

=

zz

yyb EI

EID

0

0][

=

GA

GADs 0

0][

Tziyixiiiii wvud θθθ=

Timoshenko Beams


AE-245 26


=

xi

xibi N

NB

,

,

00000

00000][

−=

0000

0000][

,

,

ixi

ixisi NN

NNB

]00000[][ ,xiai NB =

]00000[][ ,xiti NB =

Timoshenko Beams


AE-245 27

Strain discretization

][][11 ,

,

,

, dBdBN

Nb

m

iibi

m

i zixi

yixi

xz

xy ==

=

∑∑==

θθ

θθ

][][11 ,

,

,

,dBdB

NwN

NvN

w

vs

m

iisi

m

i yiiixi

ziiixi

yx

zx ==

+−

=

+−

∑∑==

θθ

θθ

][][11

,, dBdBuNu a

m

iiai

m

iixix === ∑∑

==

][][11

,, dBdBN t

m

iiti

m

ixixixx === ∑∑

==

θθ

Timoshenko Beams


AE-245 28

Load discretization

=

=

∑= ][

][

][

1 dN

dN

dN

wN

vN

uN

w

v

u

w

v

um

iii

ii

ii

δδδ

δδδ

δδδ

Timoshenko Beams


AE-245 29

Substitution into the weak form equation yields the element stiffness matrix

∫=h

bbT

bbe dxBDB0

]][[][k ∫=h

ssT

sse dxBDB0

]][[][k

∫=h

aT

aae dxBEAB0

])[(][k ∫=h

txxT

tte dxBGJB0

])[(][k

teaesebee kkkkk +++=

Timoshenko Beams


AE-245 30

Substitution into the weak form equation yields the element load vector

+++

+++

+++= ∫

Tzzz

Tyyy

Txxx

Twzz

Tvyy

Tuxx

hT

wzT

vyT

uxe

hNMhNMhNM

hNQhNQhNQ

dxNqNqNq

)]([)]([)]([

)]([)]([)]([

)][][][(0

θθθ

f

Timoshenko Beams


AE-245 31

Use one global reference system

x

z

X

Z

Y

y

Global system XYZ

Local system xyz

U V

Wv

u

w

=

W

V

U

w

v

u

t

ΘΘΘ

=

x

x

x

z

y

x

t

θθθ

Timoshenko Beams


AE-245 32

Transformation of element arrays

Ttt =−1

tktk eT

e =ˆe

Te ftf =ˆ

Timoshenko Beams


AE-245 33

Shear locking is present if the same interpolation functions are used for displacement and rotation.

x

xw

10

10

ββψαα

+=+=

xw xxz 101, )( ββαψγ ++=+=

2 point Gaussian integration is exact

This imposes two constrains

Timoshenko Beams


AE-245 34

Degrees of freedom

...

Degrees of freedom per elementn

n )1(2 +

for a very fine mesh (n→∞) 2)1(2 ≈+

n

n

n elements

Timoshenko Beams


AE-245 35

2 point Gaussian quadrature: 2 integration points, 1 constraint per point ⇒ 2 constraints per element = 2 degrees of freedom per element

1 point Gaussian quadrature: 1 integration point, 1 constraint per point ⇒ 1 constraint per element < 2 degrees of freedom per element

Timoshenko Beams


AE-245 36

Timoshenko BeamsReduced and selective integration

321reduced

integration

432complete

integration

cubicquadraticlinearshape

functions


AE-245 37

Example: cantilevered beam

0.512×10-30.9990.9810.99916

0.128×10-30.9960.9270.9968

0.320×10-40.9840.7620.9854

0.800×10-40.9380.4450.9402

0.200×10-40.7500.0420.7621

Thin beam: two

point

Thin beam:

one point

Thick beam:

two point

Thick beam:

one point

number of

elements



AE-245 38

Exercise: Consider a 2D beam in bending. Neglect axial and torsional effects and assume linear shape functions.

ψ2ψ1

w2w1

h



AE-245 39

(a) Derive the bending stiffness matrix:

−

−=

1010

0000

1010

0000

h

EIbk



AE-245 40

(b) Derive the shear stiffness matrix using 1 and 2 Gauss points

−−

−−−−

=

4/2/4/2/

2/12/1

4/2/4/2/

2/12/1

22

22

hhhh

hh

hhhh

hh

h

GAsk

−−

−−−−

=

3/2/6/2/

2/12/1

6/2/3/2/

2/12/1

22

22

hhhh

hh

hhhh

hh

h

GAsk

1 point

2 points



AE-245 41

(c) Show that kS-(1 point) has rank 1 and kS-(2 points) has rank 2 and



AE-245 42

Plates


AE-245 43

Assumptions

• Domain: Ω = (x,y,z)∈ℜ3| −t/2 ≤ z ≤ t/2, (x,y)∈ℜ2

• σzz = 0

• u(x,y,z) = u(x,y) + zψx(x,y)

• v(x,y,z) = v(x,y) + zψy(x,y)

• w(x,y,z) = w(x,y)

Mindlin Plates


AE-245 44

• The thickness t may be a function of x, y

• σzz = 0 is the plane stress assumption

• Plane sections remain plane

x

y

z

ψx

ψyCarefully check sign convention

Mindlin Plates


AE-245 45

x

zw,x

γxzγxz = w,x + ψx

w

−ψx

Mindlin Plates


AE-245 46

Strain -displacement relations

)(2

)(2

1

2

1,,,,

,,

,,

xyyxxyxy

yyyyy

xxxxx

zvu

x

v

y

u

zvy

v

zux

u

ψψε

ψε

ψε

+++=

∂∂+

∂∂=

+=∂∂=

+=∂∂=

)(2

1

2

1

)(2

1

2

1

0

,

,

,

yyyz

xxxz

zzz

wy

w

z

v

wx

w

z

u

wz

w

ψε

ψε

ε

+=

∂∂+

∂∂=

+=

∂∂+

∂∂=

==∂∂=

Mindlin Plates


AE-245 47

u, v in-plane displacements

w transverse displacement

ψα rotation angle

U, V, W prescribed displacements

Ψx, Ψy prescribed rotations

Mindlin Plates


AE-245 48

Mindlin Plates

Plate infinitesimal element: stress distributions

τxy

zdz

t/2

t/2

dydx

n

τyx

z

yx

σyyσxxz

dz

t/2

t/2

dydx

n

z

yx

τxz τyz


AE-245 49

∫

∫

∫

−

−

−

=

=

=

2/

2/

2/

2/

2/

2/

t

t

xyxy

t

t

yyyy

t

t

xxxx

dzzM

dzzM

dzzM

τ

σ

σ

Moments Shear forces

∫

∫

−

−

=

=

2/

2/

2/

2/

t

t

yzyy

t

t

xzxx

dzQ

dzQ

τ

τ

Mindlin Plates

∫

∫

∫

−

−

−

=

=

=

2/

2/

2/

2/

2/

2/

t

t

xyxy

t

t

yyyy

t

t

xxxx

dzN

dzN

dzN

τ

σ

σ

Membrane forces


AE-245 50

Mindlin Plates

Plate infinitesimal element: internal membrane forc es

t/2

t/2

dydx

dxx

NN xy

xy ∂∂

+

dyy

NN yy

yy ∂∂

+dx

x

NN xx

xx ∂∂+

xxNyyN

∫

∫

∫

−

−

−

==

=

=

2/

2/

2/

2/

2/

2/

t

t

xyyxxy

t

t

yyyy

t

t

xxxx

dzNN

dzN

dzN

τ

σ

σ

z

yx

xyNyxN

dyy

NN yx

yx ∂∂

+


AE-245 51

Mindlin Plates

Force equilibrium along x

Force equilibrium along y

0=−

∂∂

++−

∂∂+ dxNdxdy

y

NNdyNdydx

x

NN yx

yxyxxx

xxxx

0=∂

∂+

∂∂

y

N

x

N yxxx

0=−

∂∂

++−

∂∂

+ dyNdydxx

NNdxNdxdy

y

NN xy

xyxyyy

yyyy

0=∂

∂+

∂∂

x

N

y

N xyyy


AE-245 52

Mindlin Plates

Plate infinitesimal element: internal forces

t/2

t/2

dydx q

dyy

MM yx

yx ∂∂

+dx

x

MM xy

xy ∂∂

+

dyy

MM yy

yy ∂∂

+dx

x

MM xx

xx ∂∂+

dyy

QQ yy

yy ∂∂

+dx

x

QQ xx

xx ∂∂+

xxQ

xxM

yyQ

yyM

xyM

yxM

∫

∫

∫

∫

∫

−

−

−

−

−

=

=

==

=

=

2/

2/

2/

2/

2/

2/

2/

2/

2/

2/

t

t

yzyy

t

t

xzxx

t

t

xyyxxy

t

t

yyyy

t

t

xxxx

dzQ

dzQ

zdzMM

zdzM

zdzM

τ

τ

τ

σ

σ

z

yx


AE-245 53

Mindlin Plates

0=+−

∂∂

++−

∂∂+ qdxdydxQdxdy

y

QQdyQdydx

x

QQ yy

yyyyxx

xxxx

Force equilibrium along z

0=+∂

∂+

∂∂

qy

Q

x

Q yyxx


AE-245 54

Mindlin Plates

Moment equilibrium along x

Moment equilibrium along y

02

)(

2

)(

2

)( 222

=+−

∂∂++

∂∂

+

+

∂∂

+−+

∂∂

+−

dyqdx

dyQ

dydx

y

QQdxdydy

y

QQ

dxdyy

MMdxMdydx

x

MMdyM

xxxx

xxyy

yy

yyyyyy

xyxyxy

0=−∂

∂+

∂∂

yyyyxy Q

y

M

x

M

0=−∂

∂+∂

∂xx

xxxy Qx

M

y

M


AE-245 55


yxWVU ΘΘ ,,,,

Prescribed transverse force

per unit area:

q

Mindlin Plates


AE-245 56

QPrescribed boundary shear force:

x

yQ

Qyy

Qxxy

x

Mindlin Plates

yyyxxx nQnQQ +=

n


AE-245 57

Prescribed boundary membrane forces:

Mindlin Plates

yyxx NN ,

x

yy

x

Nxy

Nxy

Nxx

Nyy

n

Nxx

Nyy

yyyyyxxy

xxyxyxxx

NnNnN

NnNnN

=+

=+


AE-245 58

yyxx MM ,Prescribed boundary moment:

x

yy

x

Mxx

Myy

Mxy

Mxy

n

Myy

Mxx

Mindlin Plates

yyyyyxxy

xxyxyxxx

MnMnM

MnMnM

=+

=+


AE-245 59

Strain -displacement relations (engineering)

Mindlin Plates

)()( ,,,,

,,

,,

xyyxxyxy

yyyyy

xxxxx

zvu

zv

zu

ψψγψεψε

+++=

+=+=

yyyz

xxxz

zzz

w

w

w

ψγψγ

ε

+=+===

,

,

, 0


AE-245 60

Constitutive relations

=

xy

yy

xx

b

xy

yy

xx

Q

γεε

τσσ

][

=

yz

xz

syz

xzQ

γγ

ττ

][

Tbb QQ ][][ =

Mindlin Plates

Tss QQ ][][ =


AE-245 61

Strong form

boundary conditions

Mindlin Plates

0

0

0

0

0

,,

,,

,,

,,

,,

=−+

=−+

=++

=+

=+

yyxxyyyy

xxyxyxxx

yyyxxx

yyyxxy

yxyxxx

QMM

QMM

qQQ

NN

NN

yxWVU ΘΘ ,,,,


yyxxyyxx MMQNN ,,,,

Prescribed forces and moments


AE-245 62

Weak form

δu, δv, δw, δψx, δψy satisfy geometric homogeneous boundary conditions

Mindlin Plates

0)(

)()(

)()(

,,

,,,,

,,,,

=Ω−+

+Ω−++Ω−+

+Ω++Ω+

∫

∫∫

∫∫

Ω

ΩΩ

ΩΩ

wdqQQ

dQMMdQMM

vdNNudNN

yyyxxx

yyyxxyyyyxxxyxyxxx

yyyxxyyxyxxx

δ

δψδψ

δδ


AE-245 63

Green’s theorem

∫∫ΓΩ

Γ⋅=Ω

∂∂+

∂∂

dwQQdwQy

wQx yyxxyyxx nδδδ ),()()(

∫

∫

Γ

Ω

Γ⋅++

=Ω

+

∂∂++

∂∂

dMMMM

dMMy

MMx

xxyyyyyxyxxx

xxyyyyyxyxxx

n),(

)()(

δψδψδψδψ

δψδψδψδψ

Mindlin Plates

∫∫

∫

ΓΓ

Ω

Γ⋅+Γ⋅

=Ω

+

∂∂++

∂∂

dvNNduNN

dvNuNy

vNuNx

yyxyxyxx

yyxyxyxx

nn δδ

δδδδ

),(),(

)()(


AE-245 64

Using Green ’s theorem

0)(

)()]()([

)]([

)]([

,,

,,,,

,,,,

=Γ+−Γ−Ω−

−Γ+−Ω+++

+Ω+++

+Ω+++

∫∫∫

∫∫

∫

∫

ΓΓΩ

ΓΩ

Ω

Ω

dMMdwQwdq

dvNuNdwQwQ

dMMM

dvuNvNuN

MQ

N

yyyxxx

yyxxyyyyxxxx

xyyxxyyyyyxxxx

xyxyyyyxxx

δψδψδδ

δδδψδδψδ

δψδψδψδψ

δδδδ

Mindlin Plates


AE-245 65

In-plane, bending and shear stiffness:

][12

][][32/

2/

2b

t

t

b Qt

dzzQD == ∫−

][][][2/

2/

s

t

t

ss QtdzQA == ∫−

(provided [Qb] is independent of z)

(provided [Qs] is independent of z)

Mindlin Plates

][][][2/

2/

b

t

t

b QtdzQA == ∫−

(provided [Qb] is independent of z)


AE-245 66

+=

=

=

∫∫−−

xyyx

yy

xxt

txy

yy

xx

b

t

txy

yy

xx

xy

yy

xx

DzdzQzdz

M

M

M

,,

,

,2/

2/

2/

2/

][][

ψψψψ

γεε

τσσ

++

=

=

=

∫∫−− yy

xx

s

t

t yz

xz

s

t

t yz

xz

yy

xx

w

wAdzQdz

Q

Q

ψψ

γγ

ττ

,

,2/

2/

2/

2/

][][

Moment and force relations

Mindlin Plates

+=

=

=

∫∫−−

xy

y

xt

txy

yy

xx

b

t

txy

yy

xx

xy

yy

xx

vu

v

u

AdzQdz

N

N

N

,,

,

,2/

2/

2/

2/

][][

γεε

τσσ


AE-245 67

Weak form: matrix equation

0)()(

][

][][

,

,

,

,

,,

,

,

,,

,

,

,,

,

,

,,

,

,

=Γ+−Γ+−Γ

−Ω−Ω

++

++

+Ω

+

++Ω

+

+

∫∫∫

∫∫

∫∫

ΓΓΓ

ΩΩ

ΩΩ

dvNuNdMMdwQ

wdqdw

wA

w

w

dDd

vu

v

u

A

vu

v

u

NMQ

yyxxyyyxxx

yy

xx

s

T

yy

xx

xyyx

yy

xx

T

xyyx

yy

xx

xy

y

x

T

xy

y

x

δδδψδψδ

δψψ

δψδδψδ

ψψψψ

δψδψδψδψ

δδδδ

Mindlin Plates


AE-245 68

Boundary conditions

Qyyyyyxxy

Mxxyxyxxx

Myyyxxx

Nyyyyyxxy

Nxxyxyxxx

MnMnM

MnMnM

QnQnQ

NnNnN

NnNnN

Γ=+Γ=+Γ=+Γ=+Γ=+

on

on

on

on

on

Myyyxxyy

Myxyxxxx

Qyyyxxx

Nyyyxxy

Nyxyxxx

nMnM

nMnM

nQnQw

nNnNv

nNnNu

Γ−Γ=+=Γ−Γ=+=Γ−Γ=+=Γ−Γ=+=Γ−Γ=+=

on0or0

on0or0

on0or0

on0or0

on0or0

δψδψδδδ

Mindlin Plates


AE-245 69

Weak form

+=

+=

xyyx

yy

xx

xyyx

yy

xx

,,

,

,

,,

,

,

δψδψδψδψ

δκψψ

ψψ

κ

++

=

++

=yy

xx

yy

xx

w

w

w

w

δψδδψδ

δγψψ

γ,

,

,

,

Mindlin Plates

+=

+=

xy

y

x

xy

y

x

vu

v

u

vu

v

u

,,

,

,

,,

,

,

δδδδ

δεε


AE-245 70

Weak form: given prescribed displacements, rotations and forces find u, v, w, ψx, ψy ∈ S such

that for all δu, δv, δw, δψx, δψy ∈ V

0)()(

][

][][

=Γ+−Γ+

−Γ−Ω−Ω

+Ω+Ω

∫∫

∫∫∫

∫∫

ΓΓ

ΓΩΩ

ΩΩ

dNNdMM

dwQwdqdA

dDdA

NM

Q

yyyxxxyyyxxx

sT

TT

δψδψδψδψ

δδγδγ

κδκεδε

Mindlin Plates


AE-245 71

Introduce shape functions

∑=

=m

iii

h uNu1

∑=

=m

iii

h wNw1

δδ

∑=

=m

ixii

hx N

1

ψψ

∑=

=m

ixii

hx N

1

δψδψ

∑=

=m

iyii

hy N

1

ψψ

∑=

=m

iyii

hy N

1

δψδψ

Mindlin Plates

∑=

=m

iii

h vNv1

∑=

=m

iii

h wNw1

∑=

=m

iii

h uNu1

δδ ∑=

=m

iii

h vNv1

δδ


AE-245 72


Tymxmmmmyx wvuwvud ψψψψ L11111 =

=000

0000

0000

][

,,

,

,

xiyi

yi

xi

mi

NN

N

N

B

=

iyi

ixi

si NN

NNB

000

000][

,

,

Mindlin Plates

Tymxmmmmyx wvuwvud δψδψδδδδψδψδδδδ L11111 =

=

xiyi

yi

xi

bi

NN

N

N

B

,,

,

,

000

0000

0000

][


AE-245 73

Strain discretization

[ ] ][11

,,

,

,

,,

,

,

dBw

v

u

B

NN

N

N

b

m

i

yi

xi

i

i

i

bi

m

iyixixiyi

yiyi

xixi

xyyx

yy

xx

=

=

+=

+= ∑∑

==

ψψψψ

ψψ

ψψψψ

κ

[ ] ][11 ,

,

,

,dBw

v

u

BNwN

NwN

w

ws

m

i

yi

xi

i

i

i

si

m

i yiiiyi

xiiixi

yy

xx =

=

++

=

++

= ∑∑==

ψψ

ψψ

ψψ

γ

Mindlin Plates

[ ] ][11

,,

,

,

,,

,

,

dBw

v

u

B

vNuN

vN

uN

vu

v

u

m

m

i

yi

xi

i

i

i

mi

m

iixiiyi

iyi

ixi

xy

y

x

=

=

+=

+= ∑∑

==

ψψ

ε


AE-245 74

Load discretization

=

=

∑=

][

][

][

][

][

1

dN

dN

dN

dN

dN

w

v

u

Nw

v

u

y

x

w

v

u

m

i

yi

xi

i

i

i

i

y

x

δδδδδ

δψδψδδδ

δψδψδδδ

ψ

ψ

Mindlin Plates

[ ][ ][ ][ ][ ]my

mx

mw

mv

mu

NNN

NNN

NNN

NNN

NNN

00000000][

00000000][

00000000][

00000000][

00000000][

1

1

1

1

1

L

L

L

L

L

=

====

ψ

ψ


AE-245 75

Substitution into the weak form equation yields the element arrays

Γ+

+Γ++Γ+Ω=

Ω+Ω+Ω=

∫

∫∫∫

∫∫∫

Γ

ΓΓΩ

ΩΩΩ

dNMNM

dNNNNdNQdNq

dBABdBDBdBAB

Me

MeQee

eee

Tyyy

Txxx

Tvyy

Tuxx

Tw

Twe

ssT

sbT

bmT

me

)][][(

)][][(][][

]][[][]][[][]][[][

ψψ

f

k

Mindlin Plates


AE-245 76

Shear constraints and locking

When the plate is thin the out of plane shear strains γxz and γyz tend to zero.

The bending element proposed without any modification is prone to shear locking .

Mindlin PlatesShear locking


AE-245 77

Consider a four-node isoparametric quadrilateral element with its sides parallel to the global reference axes.

xyyx

xyyx

xyyxw

hy

hx

h

3210

3210

3210

γγγγψ

ββββψ

αααα

+++=

+++=

+++=



AE-245 78

Thin plate limit (t → 0)

0)(

0)(

321302,

323101,

=+++++=+=

=+++++=+=

xyyxw

xyyxw

yyyz

xxxz

γγγαγαψγββαββαψγ

× ×

××

2×2 Gaussian integration is exact

This imposes eight constrains



AE-245 79


Mesh with several elements

nx elements

ny elements

nnodes= (nx+1)(ny+1)

nelements=nxny


AE-245 80

Degrees of freedom per element

yx

yx

nn

nn )1)(1(3 ++

for a very fine mesh (nx→ ∞ and ny→ ∞)

3)1)(1(3

≈++

yx

yx

nn

nn



AE-245 81

2×2 Gaussian quadrature: 4 integration points, 2 constraints per point ⇒ 8 constraints per element > 3 degrees of freedom per element

1×1 Gaussian quadrature: 1 integration point, 2 constraints per point ⇒ 2 constraints per element < 3 degrees of freedom per element



AE-245 82

3×3 shear

4×4 bend

2×2 shear

3×3 bend

1×1 shear

2×2 bendselective

reduced int.

3×32×21×1uniform

reduced int.

bicubicbiquadraticbilinearshape

functions

Mindlin PlatesReduced and selective integration


AE-245 83

• Reduced integration is equivalent to mixed formulation

• Shear forces are also variables (Lagrange multipliers) in addition to displacement and rotations

• Situation similar to the nearly incompressible elasticity

Mindlin PlatesEquivalence with mixed methods


AE-245 84

• Reduced integration alleviates shear locking but introduces zero energy modes

• Possibility of singular global matrices is worrisome

112444number of zero energy modes

S3S2S1U3U2U1element

U=uniform, S=selective, 1=bilinear, 2=biquadratic, 3=bicubic

Mindlin PlatesRank deficiency


AE-245 85

101number of zero energy modes

S2S2U2integration scheme

LagrangeLagrangeserendipityψx, ψy-shape

functions

Lagrangeserendipityserendipityw-shape functions

LagrangeHeterosisSerendipityw, ψx, ψy

ψx, ψy

PlatesHeterosis element


AE-245 86

• Use of two interpolatory schemes is a drawback

• Start with Lagrange element and restrain the transverse displacement of the internal node to obtain seredipity shape functions

PlatesHeterosis element


AE-245 87

• Special procedure used to interpolate transverse shear strains

• Correct rank is achieved

PlatesT1 element


AE-245 88

ξ

η

1

2

34

e11

e12

e22

e21

e32

e31

e41

e42

g1

g2

g3

g4

h1

h2h3

h4

PlatesT1 element


AE-245 89

For each element side define strain component

++⋅+−=2

,2

212111

1

121

yyxx

h

wwg

ψψψψe

++⋅+−=2

,2

333221

2

232

yyxx

h

wwg

ψψψψe

PlatesT1 element


AE-245 90

For each node define shear strain vector

γb

eb1

eb2

b

γb1

γb2

gb1

gb2

a

ab

bb

bbb

bbbbb

bbbbb

bbbbb

gg

gg

gg

gg

−==

⋅=−−=

−−=

+=

2

1

21

2122

2211

2211

)1/()(

)1/()(

ee

eeγ

αααγααγ

γγ

PlatesT1 element


AE-245 91

Interpolate nodal values

∑=

=4

1iiiN γγ

PlatesT1 element


AE-245 92

1. If nodal displacements and rotations are assigned as to represent a constant transverse shear state then constant transverse shear modes are exactly represented.

2. In rectangular elements the shear strains vary linearly with x and y.

PlatesT1 element: remarks


AE-245 93

3. Implementation differs only in the computation of the element shear stiffness matrix through [Bs].

4. 2×2 Gaussian quadrature is employed to integrate all element contributions.

5. The element QUAD4 in Nastran has essential features in common with the T1 element.

PlatesT1 element: remarks


AE-245 94

ξ

η

1

2

3

e11

e12

e22

e21

e32

e31

g1

g2

g3

h1

h2

h3

PlatesT1 element: triangular version


AE-245 95

1. Three node triangles exhibits severe shear locking under normal circumstances.

2. The T1 element procedures described apply to the triangular version of the element without modifications.

3. Element TRIA3 in Nastran has similar features except that a “residual bending flexibility” is added.

PlatesT1 element: triangular version


AE-245 96

PlatesThe Discrete Kirchhoff approach

• The idea is to insist on satisfaction of zero transverse shear strains at a discrete number of points.

• Development of a DKT four node quadrilateral element


AE-245 97

Start with the 8 node serendipity shape functions to interpolate rotations

∑=

=

=8

1i yi

xii

y

xN ψ

ψψψ

ψ

16 degrees of freedom



AE-245 98

The normal component of ψ is required to vary linearly

2

)(2

)(2

)(2

)(

148

437

326

215

ψψnψn

ψψnψn

ψψnψn

ψψnψn

+⋅=⋅

+⋅=⋅

+⋅=⋅

+⋅=⋅

n

s

4 constraints



AE-245 99

Transverse displacement varying cubically only along the element edges. Degrees of freedom:

12 degrees of freedom

yx www ,,



AE-245 100

Kirchhoff constraints imposed

8 constraints

00 ,, =+=+ yyxx ww ψψ corner nodes

0, =++ yyxxs ssw ψψ midside nodes

4 constraints



AE-245 101

Total number of degrees of freedom

16+12 degrees of freedom

4+8+4 constraints

+

=

12 degrees of freedom (1 disp . + 2 rot. per node)



AE-245 102

The transverse shear stiffness matrix is ignored and bending stiffness is defined by the interpolation functions

Tyxyx

iiyiy

iixix

wwd

dN

dN

121212111

12

1

12

1

ψψψψ

ψ

ψ

L=

=

=

∑

∑

=

=



AE-245 103

1. The tangential shear strain components vanishes identically along the edges.

2. Exact integration requires 3 ×3 Gaussian quadrature. In practice 2 ×2 quadraturemaintains correct rank.

3. No interpolation for transverse displacement is present what leads to problems in the definition of consistent force vectors

PlatesThe DKT approach: remarks


AE-245 104

4. Convergence to thin plate solution is expected. However, for thick plates, the DKT element may yield incorrect results.

5. The triangular version of the DKT element exists and actually came first.

PlatesThe DKT approach: remarks

Documents

FINITE ELEMENTS I - Instituto Tecnológico de Aeronáuticaarfaria/AE245_08.pdf · Instituto Tecnológico de Aeronáutica AE-245 3 • Euler-Bernoulli beam model and Kirchhoff plate