Lecture Notes: Introduction to Finite Element Method |
Chapter 6. Solid Elements for 3-D Problems |
III. Typical 3-D Solid Elements
Tetrahedron:
linear (4 nodes) |
quadratic (10 nodes) |
Hexahedron (brick):
linear (8 nodes) |
quadratic (20 nodes) |
Penta:
linear (6 nodes) |
quadratic (15 nodes) |
Avoid using the linear (4-node) tetrahedron element in 3-D stress analysis (Inaccurate! However, it is OK for static deformation or vibration analysis).
© 1997-2002 Yijun Liu, University of Cincinnati |
144 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 6. Solid Elements for 3-D Problems |
Element Formulation:
Linear Hexahedron Element
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y 8 |
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mapping (xyz↔ξηζ) |
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(-1≤ ξ,η,ζ ≤ 1) |
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(-1,1,-1) 4 |
3 (1,1,-1) |
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(-1,1,1) 8 |
7 (1,1,1) |
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ξ |
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(-1,-1,-1) 1 |
2 (1,-1,-1) |
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(-1,-1,1) 5 |
6 (1,-1,1) |
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ζ
Displacement field in the element:
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u =∑Ni ui , |
v =∑Ni vi |
, w =∑Ni wi |
(11) |
i=1 |
i=1 |
1i=1 |
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© 1997-2002 Yijun Liu, University of Cincinnati |
145 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 6. |
Solid Elements for 3-D Problems |
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Shape functions: |
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N1 (ξ,η,ζ ) = |
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(1−ξ)(1−η)(1−ζ ) , |
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N2 (ξ,η,ζ ) = |
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(1+ξ)(1−η)(1−ζ ) , |
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N3 (ξ,η,ζ ) = |
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(1+ξ)(1+η)(1−ζ ) , |
(12) |
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M |
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N8 (ξ,η,ζ ) = |
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(1−ξ)(1+η)(1+ζ ) . |
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Note that we have the following relations for the shape functions:
Ni (ξ j ,ηj ,ζ j ) =δij , i, j =1,2,L,8.
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∑Ni (ξ,η,ζ ) =1.
i =1
Coordinate Transformation (Mapping):
8
x =∑Ni xi
i =1
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, y =∑Ni yi |
, z =∑Ni zi . |
(13) |
i =1 |
i =1 |
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The same shape functions are used as for the displacement field.
Isoparametric element.
© 1997-2002 Yijun Liu, University of Cincinnati |
146 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 6. Solid Elements for 3-D Problems |
Jacobian Matrix:
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∂x |
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∂η |
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∂u |
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∂ζ |
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∂y
∂ξ ∂y
∂η ∂y
∂ζ
≡
∂z |
∂u |
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∂ξ |
∂x |
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∂z |
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∂u |
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∂y |
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∂η |
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∂z |
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∂u |
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∂ζ |
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∂z |
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J Jacobian matrix
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∂u |
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∂x |
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−1 |
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∂u |
= J |
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∂y |
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∂u |
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∂z |
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and
∂v |
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∂v |
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∂x |
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∂ξ |
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−1 |
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∂v |
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∂v |
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∂y |
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∂η |
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∂v |
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∂v |
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∂ζ |
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∂z |
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∂u∂ξ∂u∂η∂u∂ζ
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∂u |
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∂N |
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= ∑ |
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ui |
,etc. |
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∂ξ |
∂ξ |
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i =1 |
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(15)
also for w.
© 1997-2002 Yijun Liu, University of Cincinnati |
147 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 6. Solid Elements for 3-D Problems |
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ε x |
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ε |
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ε |
y |
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ε= |
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γ xy |
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γ |
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yz |
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γ zx |
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∂u |
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∂x |
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∂v |
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∂y |
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∂w |
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∂x |
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=Luse (15) |
= B d |
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∂u |
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where d is the nodal displacement vector, |
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i.e., |
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ε= Bd |
(16) |
(6×1) (6×24)×(24×1)
© 1997-2002 Yijun Liu, University of Cincinnati |
148 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 6. Solid Elements for 3-D Problems |
Strain energy,
U = |
1 ∫σT εdV = |
1 ∫(Eε)T εdV |
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2 V |
2 V |
=1 ∫εT EεdV
2 V
= |
1 |
d |
T |
T |
EB dV |
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∫B |
d |
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Element stiffness matrix, k =∫BT EB dV
V
(24×24) (24×6)×(6×6)×(6×24)
In ξηζ coordinates:
dV =(det J ) dξ dηdζ
1 1 1
k = ∫ ∫ ∫BT E B (det J) dξ dη dζ
−1 −1 −1
( Numerical integration)
(17)
(18)
(19)
(20)
• 3-D elements usually do not use rotational DOFs.
© 1997-2002 Yijun Liu, University of Cincinnati |
149 |
Lecture Notes: Introduction to Finite Element Method |
Chapter 6. Solid Elements for 3-D Problems |
Treatment of distributed loads:
Distributed loads Nodal forces
pA/3 
pA/12
p
Area =A |
Nodal forces for 20-node |
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Hexahedron |
Stresses:
σ=Eε= EBd
Principal stresses:
σ1 ,σ2 ,σ3 .
von Mises stress:
σe =σVM = |
1 |
(σ1 −σ2 )2 + (σ2 −σ3 )2 + (σ3 −σ1 )2 . |
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Stresses are evaluated at selected points (including nodes) on each element. Averaging (around a node, for example) may be employed to smooth the field.
Examples: …
© 1997-2002 Yijun Liu, University of Cincinnati |
150 |