FINITE ELEMENTS I - Instituto Tecnológico de Aeronáuticaarfaria/AE245_08.pdf · Instituto...
Transcript of FINITE ELEMENTS I - Instituto Tecnológico de Aeronáuticaarfaria/AE245_08.pdf · Instituto...
Instituto Tecnológico de Aeronáutica
AE-245 1
FINITE ELEMENTS I
Class notes
Instituto Tecnológico de Aeronáutica
AE-245 2
8. Beams and Plates
Instituto Tecnológico de Aeronáutica
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• 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
Instituto Tecnológico de Aeronáutica
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Beams
Instituto Tecnológico de Aeronáutica
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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
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2. σyy = σzz = τyz = 0
Assumption used in the constitutive equation to eliminate εyy, εzz and γyz.
Timoshenko Beams
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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Strain -displacement relations
yxxxxzxz
zxxxxyxy
xyxzxxxx
ywwu
zvvu
zyuu
θθγθθγ
θθε
++=+=
−−=+=
+−==
,,,,
,,,,
,,,,
Timoshenko Beams
Instituto Tecnológico de Aeronáutica
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∫
∫
∫
−=
=
−=
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
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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zzyyxx QQQ ,,Prescribed boundary forces:
Qyy
y
x
z
Qzz
Qxx
Timoshenko Beams
Instituto Tecnológico de Aeronáutica
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zzyyxx MMM ,,Prescribed boundary moment:
Myy
y
x
z
Mzz
Mxx
Timoshenko Beams
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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Equilibrium equation: moment about x
y
x
z
Mxx + dMxx
Timoshenko Beams
Mxx
dx 0, =xxxM
Instituto Tecnológico de Aeronáutica
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Equilibrium equation: moment about y
y
x
z
Timoshenko Beams
Qzz + dQzzMyy + dMyy
Qzz
Myy
dx
qzz0, =− zzxyy QM
Instituto Tecnológico de Aeronáutica
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Equilibrium equation: moment about z
y
x
z
Timoshenko Beams
Mzz + dMzzQyy + dQyy
Mzz
Qyy
dx
qyy
0, =+ yyxzz QM
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Strong form
Timoshenko Beams
0
0
0
,
,
,
=+
=+=+
zxzz
yxyy
xxxx
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
Instituto Tecnológico de Aeronáutica
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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
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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
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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
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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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
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Define vectors and matrices
=
xi
xibi N
NB
,
,
00000
00000][
−=
0000
0000][
,
,
ixi
ixisi NN
NNB
]00000[][ ,xiai NB =
]00000[][ ,xiti NB =
Timoshenko Beams
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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Load discretization
=
=
∑= ][
][
][
1 dN
dN
dN
wN
vN
uN
w
v
u
w
v
um
iii
ii
ii
δδδ
δδδ
δδδ
Timoshenko Beams
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 32
Transformation of element arrays
Ttt =−1
tktk eT
e =ˆe
Te ftf =ˆ
Timoshenko Beams
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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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
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Timoshenko BeamsReduced and selective integration
321reduced
integration
432complete
integration
cubicquadraticlinearshape
functions
Instituto Tecnológico de Aeronáutica
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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
Timoshenko BeamsReduced and selective integration
Instituto Tecnológico de Aeronáutica
AE-245 38
Exercise: Consider a 2D beam in bending. Neglect axial and torsional effects and assume linear shape functions.
ψ2ψ1
w2w1
h
Timoshenko BeamsReduced and selective integration
Instituto Tecnológico de Aeronáutica
AE-245 39
(a) Derive the bending stiffness matrix:
−
−=
1010
0000
1010
0000
h
EIbk
Timoshenko BeamsReduced and selective integration
Instituto Tecnológico de Aeronáutica
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
Timoshenko BeamsReduced and selective integration
Instituto Tecnológico de Aeronáutica
AE-245 41
(c) Show that kS-(1 point) has rank 1 and kS-(2 points) has rank 2 and
Timoshenko BeamsReduced and selective integration
Instituto Tecnológico de Aeronáutica
AE-245 42
Plates
Instituto Tecnológico de Aeronáutica
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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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 45
x
zw,x
γxzγxz = w,x + ψx
w
−ψx
Mindlin Plates
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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 ∂∂
+
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 53
Mindlin Plates
0=+−
∂∂
++−
∂∂+ qdxdydxQdxdy
y
QQdyQdydx
x
QQ yy
yyyyxx
xxxx
Force equilibrium along z
0=+∂
∂+
∂∂
qy
Q
x
Q yyxx
Instituto Tecnológico de Aeronáutica
AE-245 54
Mindlin Plates
Moment equilibrium along x
Moment equilibrium along y
02
)(
2
)(
2
)( 222
=+−
∂∂++
∂∂
+
+
∂∂
+−+
∂∂
+−
dyqdx
dyQ
dydx
y
QQdxdydy
y
dxdyy
MMdxMdydx
x
MMdyM
xxxx
xxyy
yy
yyyyyy
xyxyxy
0=−∂
∂+
∂∂
yyyyxy Q
y
M
x
M
0=−∂
∂+∂
∂xx
xxxy Qx
M
y
M
Instituto Tecnológico de Aeronáutica
AE-245 55
Prescribed displacements and rotations
yxWVU ΘΘ ,,,,
Prescribed transverse force
per unit area:
q
Mindlin Plates
Instituto Tecnológico de Aeronáutica
AE-245 56
QPrescribed boundary shear force:
x
yQ
Qyy
Qxxy
x
Mindlin Plates
yyyxxx nQnQQ +=
n
Instituto Tecnológico de Aeronáutica
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
=+
=+
Instituto Tecnológico de Aeronáutica
AE-245 58
yyxx MM ,Prescribed boundary moment:
x
yy
x
Mxx
Myy
Mxy
Mxy
n
Myy
Mxx
Mindlin Plates
yyyyyxxy
xxyxyxxx
MnMnM
MnMnM
=+
=+
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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 ][][ =
Instituto Tecnológico de Aeronáutica
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 ,,,,
Prescribed displacements and rotations
yyxxyyxx MMQNN ,,,,
Prescribed forces and moments
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
),(),(
)()(
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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)
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 72
Define vectors and matrices
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
][
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
+++=
+++=
+++=
Mindlin PlatesShear locking
Instituto Tecnológico de Aeronáutica
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
Mindlin PlatesShear locking
Instituto Tecnológico de Aeronáutica
AE-245 79
Mindlin PlatesShear locking
Mesh with several elements
nx elements
ny elements
nnodes= (nx+1)(ny+1)
nelements=nxny
Instituto Tecnológico de Aeronáutica
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
Mindlin PlatesShear locking
Instituto Tecnológico de Aeronáutica
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
Mindlin PlatesShear locking
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 85
101number of zero energy modes
S2S2U2integration scheme
LagrangeLagrangeserendipityψx, ψy-shape
functions
Lagrangeserendipityserendipityw-shape functions
LagrangeHeterosisSerendipityw, ψx, ψy
ψx, ψy
PlatesHeterosis element
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 87
• Special procedure used to interpolate transverse shear strains
• Correct rank is achieved
PlatesT1 element
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 91
Interpolate nodal values
∑=
=4
1iiiN γγ
PlatesT1 element
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
AE-245 94
ξ
η
1
2
3
e11
e12
e22
e21
e32
e31
g1
g2
g3
h1
h2
h3
PlatesT1 element: triangular version
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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
PlatesThe Discrete Kirchhoff approach
Instituto Tecnológico de Aeronáutica
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
PlatesThe Discrete Kirchhoff approach
Instituto Tecnológico de Aeronáutica
AE-245 99
Transverse displacement varying cubically only along the element edges. Degrees of freedom:
12 degrees of freedom
yx www ,,
PlatesThe Discrete Kirchhoff approach
Instituto Tecnológico de Aeronáutica
AE-245 100
Kirchhoff constraints imposed
8 constraints
00 ,, =+=+ yyxx ww ψψ corner nodes
0, =++ yyxxs ssw ψψ midside nodes
4 constraints
PlatesThe Discrete Kirchhoff approach
Instituto Tecnológico de Aeronáutica
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)
PlatesThe Discrete Kirchhoff approach
Instituto Tecnológico de Aeronáutica
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=
=
=
∑
∑
=
=
PlatesThe Discrete Kirchhoff approach
Instituto Tecnológico de Aeronáutica
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
Instituto Tecnológico de Aeronáutica
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