MMJ 1113 FINITE ELEMENT METHOD FEM in Elasticity Task: To derive the element stiffness matrix for a 3-node triangular element under plane stress / plane strain condition in elasticity. Requirements: Constitutive equations (Generalized stress-strain relations) For variational approach - Examine the functional (based on potential energy) Minimize the functional - yield finite element equations: [k]{δ} = {f} Finite Element in Elasticity
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MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Task:To derive the element stiffness matrix for a 3-node triangular element under plane stress / plane strain condition in elasticity.
The linear interpolation model satisfies both compatibility andcompleteness requirements
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Strain–displacement relationships
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
∂Ν∂
=
∂∂
+∂∂∂∂∂∂
=
=
3
3
2
2
1
1
332211
321
321
vuvuvu
xyxyxy
yyy
xxx
yu
xvyvxu
xy
y
x
γεε
ε
[ ] δε Β=
0 0
0
0
00
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Derive the element equations using the minimum potential energy principle.General element equations2-dimensional (plane) elasticity problemsApply these equations for case of plane stress, plane strain and
axisymmetric solids.
The functional in the variational principle is the potential energy of a 2-D elastic body acted by surface and body forces.
The area, A, is divided into M discrete elements.
( ) ( ) ( )∑=
=M
e
e vuvu1
,, ππ
T*
F*
C1
x, u
y, v
t(x,y)= thickness
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
To ensure convergence as element size decreases, the chosen interpolation functions must satisfy compatibility and completeness requirements.
Examine the potential energy equation:
CompatibilityOnly first–order derivatives of the displacement appear in the integrand of the functional.Thus the interpolation functions must ensure continuous displacement at element interfaces.
i.e. C0 continuity in displacement at interfaces.CompletenessThe interpolation functions must ensure that rigid body displacement (uniform displacement) and constant-strain states (uniform first derivatives of displacement) are represented in the limit as element size is reduced.
i.e. C1 continuity of displacement within elements.
( ) [ ] [ ][ ] [ ] [ ] [ ]
[ ] dSTtdA
tdACBtdABCBvu
cA
oT
AA
T
δδ
εδδδπ
~ ~F-
~221 ~~
21,
1
**
*
∫∫∫
∫∫∫∫−
−=
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Polynomial containing at least one constant and the linear terms can serve as interpolation functions satisfying both compatibility and completeness requirements.
For an element (e) with r number of nodes:
( ) ( )( )
( )
( )
( )
( )
( ) [ ] ( )
ntsdisplaceme nodal
~
N
,
,N
,,~
1
1i
↓
=
=
=
=
∑
∑
=
=
ee
e
r
iii
r
ii
ee
N
vNu
vyxN
uyx
yxvyxu
δδ
δ
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Minimum Potential Energy Principle
A variational principle that hold throughout the system.Potential Energy = strain energy – work by external force
Strain energy of linear elastic body
Initial strain terms
( ) ( ) ( )wvuWwvuUwvu ,,,,,, −=Π
[ ]
[ ] [ ][ ] dv~ BC~21
dv 21
dv 21
T δδ
εε
σε
B∫∫∫
∫∫∫
∫∫∫
Ω
Ω
Ω
=
=
=
C
U
[ ]
[ ] [ ] dv C~221
dv 221U
0T
0o
εδ
εε
B∫∫∫
∫∫∫
Ω
Ω
=
= C
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Work done by external force
∫∫+∫∫∫= Ω 1
~~21
S dSTdVFW δδ
Body Force Traction
ZYXF = zyx TTTT =
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
The potential energy functional
The equilibrium displacement field (u, v, w) minimizes П and satisfies all the boundary conditions.
[ ] [ ][ ] [ ] [ ] 1s
dS ~T-dV ~F-dV ~2~~21
1
δδεδδδ ∫∫∫∫∫∫∫∫ Β−ΒΒ=ΠΩΩ
oTT CC
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Theory of Minimum Potential Energy
The displacement field (u , v) which satisfies the equilibrium, and the conditions at the boundary surface is the one that minimizes the potential energy.
( ) ( )( )
( )( )( ) ( )
0,
0,,
11
1
=∑∂Π∂
+∑∂Π∂
=Π
=∑ Π=Π
==
Μ
=
i
r
i i
e
i
r
i i
ee
e
e
vv
uu
vu
vuvu
δδδ
δδvariation
( ) ( )r1,2,.....,i ; 0 ==
∂∂
=∂∂
i
e
i
e
vuππ
Since δui , δvi are independent variations and not necessarily zero, thus
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Look at typical node q of the element, we have
The form of this force-displacement relation for node q is
Additional load vector, if preset, can be added to the resultant external load vector Fq.Example: initial force vector due to initial strains from non-uniform temperature.
( )
( )
[ ] ( )
( )
[ ]( )[ ] ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
( )
0**
0
1
=Ν−Ν−ΒΒ
=
∂∂∂∂
∫∫∫∫∫eee
e
s
eq
eqq
eeq
Aq
eeqep
e
A
Tq
q
eq
e
dSTdApdAtC
v
u
δ
π
π
[ ] elementboundary for -loading surface todue vector force nodal F
vectorforcebody nodal
T
q
=
=
=+=
qB
qqT
qB
qqp
F
FFFk δ
[ ] ( )
( )
[ ]( ) [ ] ( ) ( ) ( )eeepo
e
A
Tq dAtC
e
e *ε∫∫ Β
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Plane stress versus plane strain
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Plane stress condition - small displacement
( )
( )
( )332211332211
332211
332211
21
21
21
ucucucvbvbvbyu
xv
vcvcvcyv
ubububxu
xy
y
x
+++++∆
=∂∂
+∂∂
=
++∆
=∂∂
=
++∆
=∂∂
=
λ
ε
ε
∆=
=
3
3
2
2
1
1
332211
321
321
21
vuvuvu
bcbcbcccc
bbb
xy
y
x
γεε
ε
[ ] δε Β=
000
000
12
3
u1u2
u3
v2
v3
v1
y
x
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
[ ] εσ
γεε
νν
ν
ντσσ
C
xy
y
x
xy
y
x
=
−−Ε
=
2100
0101
1 2
Plane stress conditionHomogeneous and isotropic solids
Stress-strain relations
( ) xyxy
yxy
yxx
γν
τ
εν
εν
νσ
εν
νεν
σ
+Ε
=
−Ε
+−Ε
=
−Ε
+−Ε
=
12
11
11
22
22
[ ] ( )( )
lyrespective ratio, sPoisson' and modulus sYoung' are and E where22100
0101
211
conditionstrain PlaneFor
ν
γεε
ννν
νν
νν
−−
−
−+Ε
=
xy
y
x
C
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Example:For a plane stress case, and a typical node, i
If the thickness of the element is uniform (t(e) constant)
If the thickness of the element varies, estimate
Using table 5.2 (text book)
[ ] [ ]( )( )[ ]( )[ ]( ) ( ) ( ) 1,2,3ji,
eA== ∫∫ eeeeeT
i
ij dAtBCBk
( ) ( ) ( ) ( )332211 NtNtNtt eeee ++≈
( )( )
( ) ( ) ∆=∫∫ .ee
A
e tdAte
( )( )
( ) ( ) ( ) ( )( )eeee
A
e tttdAte 3213
++∆
=∫∫
( )( )
( )eA
e
ii
i
i
ii
ii dAtbcc
b
bccbE
e∫∫
−
−∆
= 00
2100
0101
014
122 ν
νν
ν 0
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
( )
( ) ( )
( )
∆
=
= ∫∫
*
*e
ee*
*
A
3t
dA te
i
i
i
ii
iB
YX
YXNF
dS *
*
1
= ∫yi
xi
ci
iT T
TNF
Force vector terms, F(e)
Body forces
Surface traction
k(e)
j
i
Lij
s
y
x
( )
=
=
−=⇒
−=Ν
∫
*
*ij
*
*
0ij
2
*
*
0
2L
2Ls-s
dS 1
1
yi
xi
yi
xi
L
yi
xiL
ij
it
iji
TT
TT
TT
LsF
Lss
ij
ij
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Arrangement of termsdof at each node : 2No. of nodes for each element: rElement matrix : Force-displacement equations: [ ]( ) ( ) ( )eee Fk =δ
[ ]( )
[ ] [ ] [ ][ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
=
rrrr
qrqq
r
r
e
kkk
kkk
kkkkkk
k
....................
....................
....
....
21
21
22221
11211
( )
( )
=
=
=
r
qe
r
r
r
e
F
F
FF
vu
vuvu
.....
....F , ........
2
1
2
2
1
1
2
1
δ
δδ
δdiscrete nodal disp.
[k](e)
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Equivalent Nodal Force Vector
STEPS:
Calculate the total load acting on each element surface due to applied distributed loading.Split the total load proportionally to each of the node along the element edge.Sum the load vector at each nodeResolve the resultant nodal force into the desired (x-,y-) component.
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Example 1
Task: To determine the displacement (deflection) and stress field in the plate subjected to the given loadingsince t is only 0.1% of the next smallest dimension of the plate, assume plane stress condition exist.Due to symmetry (in both geometry and loading) along a vertical plane in the center of the plate, consider only one half of the region for FE discretizationChoose 3-node triangular elements (this will yield a constant strain field within each element).
3020 20 20
w=10kN/cm2
E=20x106N/cm2
υ=0.25t=3cm
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Boundary Conditions
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Continuum discretization
3-node triangular elementsTotal of 18 elements and
16 nodes.
MMJ 1113 FINITE ELEMENT METHODFEM in Elasticity
Result
Both tensile and compressive stresses occurs because plate in bending