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Introduction to Finite Element Methods - Constant Strain Triangle
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ONLY FOR INFORMATION PURPOSE
INTRODUCTION TO FINITE ELEMENT METHODS
CONSTANT STRAIN TRIANGLE
Introduction to Finite Elements
Constant Strain Triangle (CST)
Summary:
• Computation of shape functions for constant strain triangle• Properties of the shape functions • Computation of strain-displacement matrix• Computation of element stiffness matrix• Computation of nodal loads due to body forces• Computation of nodal loads due to traction• Recommendations for use• Example problems
Finite element formulation for 2D:
Step 1: Divide the body into finite elements connected to each other through special points (“nodes”)
x
y
Su
STu
v
x
px
py
Element ‘e’
3
21
4
y
xvu
1
2
3
4
u1
u2
u3
u4
v4
v3
v2
v1
=
4
4
3
3
2
2
1
1
v
u
v
u
v
u
v
u
d
44332211
44332211
vy)(x,N vy)(x,N vy)(x,N vy)(x,Ny)(x,v
u y)(x,Nu y)(x,Nu y)(x,Nu y)(x,Ny)(x,u
+++≈+++≈
=
=
4
4
3
3
2
2
1
1
4321
4321
v
u
v
u
v
u
v
u
N0N0N0N0
0N0N0N0N
y)(x,v
y)(x,uu
dNu =
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT
...... vy)(x,N
uy)(x,Ny)(x,vy)(x,u
vy)(x,N
vy)(x,N
v y)(x,N
vy)(x,Ny)(x,v
u y)(x,N
u y)(x,N
u y)(x,N
u y)(x,Ny)(x,u
11
11
xy
44
33
22
11
y
44
33
22
11
x
+∂
∂+
∂∂
≈∂
∂+∂
∂=
∂∂
+∂
∂+
∂∂
+∂
∂≈
∂∂=
∂∂
+∂
∂+
∂∂
+∂
∂≈
∂∂=
xyxy
yyyyy
xxxxx
γ
ε
ε
Approximation of the strain in element ‘e’
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
=
4
4
3
3
2
2
1
1
B
44332211
4321
4321
xy
v
u
v
u
v
u
v
u
y)(x,Ny)(x,Ny)(x,Ny)(x,N y)(x,N y)(x,Ny)(x,Ny)(x,N
y)(x,N0
y)(x,N0
y)(x,N0
y)(x,N0
0y)(x,N
0y)(x,N
0 y)(x,N
0y)(x,N
4444444444444444444 34444444444444444444 21xyxyxyxy
yyyy
xxxx
y
x
γεε
ε
dBε =
Displacement approximation in terms of shape functions
Strain approximation in terms of strain-displacement matrix
Stress approximation
Summary: For each element
Element stiffness matrix
Element nodal load vector
dNu =
dBD=σ
dBε =
∫=eV
k dVBDBT
443442143421
S
eT
b
e
f
S ST
f
V
T dSTdVXf ∫∫ += NN
Constant Strain Triangle (CST) : Simplest 2D finite element
• 3 nodes per element• 2 dofs per node (each node can move in x- and y- directions)• Hence 6 dofs per element
x
y
u3
v3
v1
u1
u2
v2
2
3
1
(x,y)
vu
(x1,y1)
(x2,y2)
(x3,y3)
166212 dNu ××× =
=
=
3
3
2
2
1
1
321
321
v
u
v
u
v
u
N0N0N0
0N0N0N
y)(x,v
y)(x,uu
The displacement approximation in terms of shape functions is
=
321
321
N0N0N0
0N0N0NN
1 1 2 2 3 3u (x,y) u u uN N N≈ + +
1 1 2 2 3 3v(x,y) v v vN N N≈ + +
Formula for the shape functions are
A
ycxbaN
A
ycxbaN
A
ycxbaN
2
2
2
3333
2222
1111
++=
++=
++=
12321312213
31213231132
23132123321
33
22
11
x1
x1
x1
det2
1
xxcyybyxyxa
xxcyybyxyxa
xxcyybyxyxa
y
y
y
triangleofareaA
−=−=−=−=−=−=−=−=−=
==
where
x
y
u3
v3
v1
u1
u2
v2
2
3
1
(x,y)
vu
(x1,y1)
(x2,y2)
(x3,y3)
Properties of the shape functions:
1. The shape functions N1, N2 and N3 are linear functions of x and y
x
y
2
3
1
1
N1
2
3
1
N2
12
3
1
1
N3
=nodesotherat
inodeat
0
''1N i
2. At every point in the domain
yy
xx
i
i
=
=
=
∑
∑
∑
=
=
=
3
1ii
3
1ii
3
1ii
N
N
1N
3. Geometric interpretation of the shape functionsAt any point P(x,y) that the shape functions are evaluated,
x
y
2
3
1P (x,y)
A1A3
A2
A
AA
AA
A
33
22
11
N
N
N
=
=
=
Approximation of the strains
xy
u
v
u v
x
y
x
B dy
y x
εεεγ
∂ ∂
∂ = = ≈ ∂ ∂ ∂ + ∂ ∂
=
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
=
332211
321
321
332211
321
321
000
000
2
1
y)(x,Ny)(x,N y)(x,N y)(x,Ny)(x,Ny)(x,N
y)(x,N0
y)(x,N0
y)(x,N0
0y)(x,N
0 y)(x,N
0y)(x,N
bcbcbc
ccc
bbb
A
xyxyxy
yyy
xxx
B
Element stresses (constant inside each element)
dBD=σ
Inside each element, all components of strain are constant: hence the name Constant Strain Triangle
IMPORTANT NOTE:1. The displacement field is continuous across element boundaries2. The strains and stresses are NOT continuous across element boundaries
Element stiffness matrix
∫=eV
k dVBDBT
AtkeV
BDBdVBDB TT == ∫ t=thickness of the elementA=surface area of the element
Since B is constant
t
A
443442143421
S
eT
b
e
f
S ST
f
V
T dSTdVXf ∫∫ += NN
Element nodal load vector
Element nodal load vector due to body forces
∫∫ ==ee A
T
V
T
bdAXtdVXf NN
=
=
∫∫∫∫∫∫
e
e
e
e
e
e
A b
A a
A b
A a
A b
A a
yb
xb
yb
xb
yb
xb
b
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
f
f
f
f
f
f
f
3
3
2
2
1
1
3
3
2
2
1
1
x
y
fb3x
fb3y
fb1y
fb1x
fb2x
fb2y
2
3
1
(x,y)
XbXa
EXAMPLE:
If X a=1 and Xb=0
=
=
=
=
∫
∫
∫
∫∫∫∫∫∫
03
03
03
0
0
0
3
2
1
3
3
2
2
1
1
3
3
2
2
1
1
tA
tA
tA
dANt
dANt
dANt
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
dAXNt
f
f
f
f
f
f
f
e
e
e
e
e
e
e
e
e
A
A
A
A b
A a
A b
A a
A b
A a
yb
xb
yb
xb
yb
xb
b
Element nodal load vector due to traction
∫=e
TS ST
SdSTf N
EXAMPLE:
x
y
fS3x
fS3y
fS1y
fS1x
2
3
1 ∫− −
=el S
along
T
SdSTtf
31 31N
Element nodal load vector due to traction
EXAMPLE:
x
y
fS3x
2
31
∫− −
=el S
along
T
SdSTtf
32 32N
fS3y
fS2x
fS2y
(2,0)
(2,2)
(0,0)
=0
1ST
tt
dyNtfex l alongS
=××
=
= ∫−
−
122
1
)1(32
2 322
0
0
3
3
2
=
=
=
y
x
y
S
S
S
f
tf
f
Similarly, compute
1
2
Recommendations for use of CST
1. Use in areas where strain gradients are small
2. Use in mesh transition areas (fine mesh to coarse mesh)
3. Avoid CST in critical areas of structures (e.g., stress concentrations, edges of holes, corners)
4. In general CSTs are not recommended for general analysis purposes as a very large number of these elements are required for reasonable accuracy.
Example
x
y
El 1
El 2
1
23
4
300 psi1000 lb
3 in
2 inThickness (t) = 0.5 inE= 30×106 psiν=0.25
(a) Compute the unknown nodal displacements.(b) Compute the stresses in the two elements.
Realize that this is a plane stress problem and therefore we need to use
psiE
D 72
10
2.100
02.38.0
08.02.3
2
100
01
01
1×
=
−−=
νν
ν
ν
Step 1: Node-element connectivity chart
ELEMENT Node 1 Node 2 Node 3 Area (sqin)
1 1 2 4 3
2 3 4 2 3
Node x y
1 3 0
2 3 2
3 0 2
4 0 0
Nodal coordinates
Step 2: Compute strain-displacement matrices for the elements
=
332211
321
321
000
000
2
1
bcbcbc
ccc
bbb
AB
Recall
123312231
213132321
xxcxxcxxc
yybyybyyb
−=−=−=−=−=−=
with
For Element #1:
1(1)
2(2)
4(3)(local numbers within brackets)
0;3;3
0;2;0
321
321
======
xxx
yyy
Hence
033
202
321
321
==−=−===
ccc
bbb
−−−
−=
200323
003030
020002
6
1)1(BTherefore
For Element #2:
−−−
−=
200323
003030
020002
6
1)2(B
Step 3: Compute element stiffness matrices
7
)1(T)1()1(T)1()1(
10
2.0
05333.0
02.02.1
3.00045.0
2.02.02.13.04.1
3.05333.02.045.05.09833.0
BDB)5.0)(3(BDB
×
−−−−
−−−
=
== Atk
u1 u2 u4 v4v2v1
7
)2(T)2()2(T)2()2(
10
2.0
05333.0
02.02.1
3.00045.0
2.02.02.13.04.1
3.05333.02.045.05.09833.0
BDB)5.0)(3(BDB
×
−−−−
−−−
=
== Atk
u3 u4 u2 v2v4v3
Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of freedom
014433 ===== vvuvuNotice that
Hence we need to calculate only a small (3x3) stiffness matrix
710
4.102.0
0983.045.0
2.045.0983.0
×
−−
=K
u1 u2v2
u1
u2
v2
Step 5: Compute consistent nodal loads
=
y
x
x
f
f
f
f
2
2
1
=
yf2
0
0
ySy ff2
10002 +−=
The consistent nodal load due to traction on the edge 3-2
lbx
dxx
dxN
tdxNf
x
x
xS y
2252
950
250
3150
)5.0)(300(
)300(
3
0
2
3
0
3
0 233
3
0 2332
−=
−=
−=
−=
−=
−=
∫
∫
∫
=
= −
= −
3 2
3232
xN =
−
lb
ffySy
1225
100022
−=
+−=Hence
Step 6: Solve the system equations to obtain the unknown nodal loads
fdK =
−=
−−
×1225
0
0
4.102.0
0983.045.0
2.045.0983.0
10
2
2
17
v
u
u
Solve to get
×−××
=
−
−
−
in
in
in
v
u
u
4
4
4
2
2
1
109084.0
101069.0
102337.0
Step 7: Compute the stresses in the elements
)1()1()1( dBD=σ
With
[ ][ ]00109084.0101069.00102337.0
d444
442211)1(
−−− ×−××=
= vuvuvuT
Calculate
psi
−−−
=1.76
1.1391
1.114)1(σ
In Element #1
)2()2()2( dBD=σ
With
[ ][ ]44
224433)2(
109084.0101069.00000
d−− ×−×=
= vuvuvuT
Calculate
psi
−=
35.363
52.28
1.114)2(σ
In Element #2
Notice that the stresses are constant in each element