Constant strain triangular

constant strain triangular which is used in analysis of triangular in finite element method with the help of shape function and natural coordinate system.

1. MANE 4240 & CIVL 4240
Introduction to Finite Elements
Constant Strain
Triangle (CST)
Prof. Suvranu De

2. Reading assignment:
Logan 6.2-6.5 + Lecture notes
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

3. Finite element formulation for 2D:
Step 1: Divide the body into finite elements connected to each
other through special points (“nodes”)
x
y
S
u
S
T
u
v
x
p
x
p
y
Element ‘e’
3
2
1
4
y
x
v
u
1
2
3
4
u
1
u
2
u
3
u
4
v
4
v
3
v
2
v
1
ï
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ï
ï
ï
þ
ï
ï
ï
ï
ï
ý
ü
ï
ï
ï
ï
ï
î
ï
ï
ï
ï
ï
í
ì
=
4
4
3
3
2
2
1
1
v
u
v
u
v
u
v
u
d

4. 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
+++»
+++»
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þ
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ý
ü
ï
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î
ï
ï
ï
ï
ï
í
ì
ú
û
ù
ê
ë
é
=
þ
ý
ü
î
í
ì
=
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,u
u
dNu=

5. TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN
EACH ELEMENT
...... v
y)(x,N
u
y)(x,Ny)(x,vy)(x,u
v
y)(x,N
v
y)(x,N
v
y)(x,N
v
y)(x,Ny)(x,v
u
y)(x,N
u
y)(x,N
u
y)(x,N
u
y)(x,Ny)(x,u
1
1
1
1
xy
4
4
3
3
2
2
1
1
y
4
4
3
3
2
2
1
1
x
+
¶
¶
+
¶
¶
»
¶
¶
+
¶
¶
=
¶
¶
+
¶
¶
+
¶
¶
+
¶
¶
»
¶
¶
=
¶
¶
+
¶
¶
+
¶
¶
+
¶
¶
»
¶
¶
=
xyxy
yyyyy
xxxxx
g
e
e
Approximation of the strain in element ‘e’

6. ï
ï
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þ
ï
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ï
ï
ý
ü
ï
ï
ï
ï
ï
î
ï
ï
ï
ï
ï
í
ì
ú
ú
ú
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ê
ê
ê
ë
é
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
=
ï
þ
ï
ý
ü
ï
î
ï
í
ì
=
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,N
0
y)(x,N
0
y)(x,N
0
y)(x,N
0
0
y)(x,N
0
y)(x,N
0
y)(x,N
0
y)(x,N
  
xyxyxyxy
yyyy
xxxx
y
x
g
e
e
e
dBε=

7. 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=s
dBε=
ò
=
e
V
k dVBDB
T

S
e
T
b
e
f
S
S
T
f
V
T
dSTdVXf òò
+= NN

8. 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
u
3
v
3
v
1
u
1
u
2
v
2
2
3
1
(x,y)
v
u
(x
1
,y
1
)
(x
2
,y
2
)
(x
3
,y
3
)

9. 166212 dNu
´´´= ï
ï
ï
ï
þ
ïï
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ï
ý
ü
ï
ï
ï
ï
î
ïï
ï
ï
í
ì
ú
û
ù
ê
ë
é
=
þ
ý
ü
î
í
ì
=
3
3
2
2
1
1
321
321
v
u
v
u
v
u
N0N0N0
0N0N0N
y)(x,v
y)(x,u
u
The displacement approximation in terms of shape functions is
ú
û
ù
ê
ë
é
=
321
321
N0N0N0
0N0N0N
N
1 1 2 2 3 3
u (x,y) u u uN N N» + +
1 1 2 2 3 3
v(x,y) v v vN N N» + +

10. Formula for the shape functions are
A
ycxba
N
A
ycxba
N
A
ycxba
N
2
2
2
333
3
222
2
111
1
++
=
++
=
++
=
12321312213
31213231132
23132123321
33
22
11
x1
x1
x1
det
2
1
xxcyybyxyxa
xxcyybyxyxa
xxcyybyxyxa
y
y
y
triangleofareaA
-=-=-=
-=-=-=
-=-=-=
ú
ú
ú
û
ù
ê
ê
ê
ë
é
==
where
x
y
u
3
v
3
v
1
u
1
u
2
v
2
2
3
1
(x,y)
v
u
(x
1
,y
1
)
(x
2
,y
2
)
(x
3
,y
3
)

11. Properties of the shape functions:
1. The shape functions N
1
, N
2
and N
3
are linear functions of x
and y
x
y
2
3
1
1
N
1
2
3
1
N
2
1
2
3
1
1
N
3
î
í
ì
=
nodesotherat
inodeat
0
''1
N
i

12. 2. At every point in the domain
yy
xx
i
i
=
=
=
å
å
å
=
=
=
3
1i
i
3
1i
i
3
1i
i
N
N
1N

13. 3. Geometric interpretation of the shape functions
At any point P(x,y) that the shape functions are evaluated,
x
y
2
3
1
P (x,y)
A
1
A
3
A
2
A
A
A
A
A
A
3
3
2
2
1
1
N
N
N
=
=
=

14. Approximation of the strains
xy
u
v
u v
x
y
x
Bd
y
y x
e
ee
g
¶ì ü
ï ï
¶
ì üï ï
¶ï ïï ï
= = »í ý í ý
¶
ï ï ï ï
î þï¶ ¶ ï
+
ï ï
¶ ¶î þ
ú
ú
ú
û
ù
ê
ê
ê
ë
é
=
ú
ú
ú
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ê
ê
ê
ë
é
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
=
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,N
0
y)(x,N
0
y)(x,N
0
0
y)(x,N
0
y)(x,N
0
y)(x,N
bcbcbc
ccc
bbb
A
xyxyxy
yyy
xxx
B

15. Element stresses (constant inside each element)
dBD=s
Inside each element, all components of strain are constant: hence
the name Constant Strain Triangle

16. IMPORTANT NOTE:
1. The displacement field is continuous across element
boundaries
2. The strains and stresses are NOT continuous across element
boundaries

17. Element stiffness matrix
ò
=
e
V
k dVBDB
T
Atk
e
V
BDBdVBDB
TT
== ò
t=thickness of the element
A=surface area of the element
Since B is constant
t
A

18. 
S
e
T
b
e
f
S
S
T
f
V
T
dSTdVXf òò
+= NN
Element nodal load vector

19. Element nodal load vector due to body forces
òò
==
ee
A
T
V
T
b
dAXtdVXf NN
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þ
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ý
ü
ï
ï
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ï
î
ï
ï
ï
ï
ï
í
ì
=
ï
ï
ï
ï
þ
ï
ï
ï
ï
ý
ü
ï
ï
ï
ï
î
ï
ï
ï
ï
í
ì
=
ò
ò
ò
ò
ò
ò
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
f
b3x
f
b3y
f
b1y
f
b1x
f
b2x
f
b2y
2
3
1
(x,y)
X
b
X
a

20. EXAMPLE:
If X
a
=1 and X
b
=0
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î
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í
ì
=
ï
ï
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þ
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ý
ü
ï
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î
ï
ï
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ï
í
ì
=
ï
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þ
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ý
ü
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î
ï
ï
ï
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ï
í
ì
=
ï
ï
ï
ï
þ
ï
ï
ï
ï
ý
ü
ï
ï
ï
ï
î
ï
ï
ï
ï
í
ì
=
ò
ò
ò
ò
ò
ò
ò
ò
ò
0
3
0
3
0
3
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

21. Element nodal load vector due to traction
ò
=
e
TS
S
T
S
dSTf N
EXAMPLE:
x
y
f
S3x
f
S3y
f
S1y
f
S1x
2
3
1 ò
- -
=
e
l
S
along
T
S
dSTtf
31 31
N

22. Element nodal load vector due to traction
EXAMPLE:
x
y
f
S3x
2
31
ò
- -
=
e
l
S
along
T
S
dSTtf
32 32
N
f
S3y
f
S2x
f
S2y
(2,0)
(2,2)
(0,0)
þ
ý
ü
î
í
ì
=
0
1
ST
tt
dyNtf
e
x
l
along
S
=´´÷
ø
ö
ç
è
æ
=
=ò
-
-
12
2
1
)1(
32
2 32
2
0
0
3
3
2
=
=
=
y
x
y
S
S
S
f
tf
f
Similarly, compute
1
2

23. 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.

24. Example
x
y
El 1
El 2
1
23
4
300 psi
1000 lb
3 in
2 in
Thickness (t) = 0.5 in
E= 30×10
6
psi
n=0.25
(a) Compute the unknown nodal displacements.
(b) Compute the stresses in the two elements.

25. Realize that this is a plane stress problem and therefore we need to use
psi
E
D
7
2
10
2.100
02.38.0
08.02.3
2
1
00
01
01
1
´
ú
ú
ú
û
ù
ê
ê
ê
ë
é
=
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ë
é
--
=
n
n
n
n
Step 1: Node-element connectivity chart
ELEMENT Node 1Node 2Node 3Area
(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

26. Step 2: Compute strain-displacement matrices for the elements
ú
ú
ú
û
ù
ê
ê
ê
ë
é
=
332211
321
321
000
000
2
1
bcbcbc
ccc
bbb
A
B
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
ú
ú
ú
û
ù
ê
ê
ê
ë
é
--
-
-
=
200323
003030
020002
6
1)1(
B
Therefore
For Element #2:
ú
ú
ú
û
ù
ê
ê
ê
ë
é
--
-
-
=
200323
003030
020002
6
1)2(
B

27. 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
´
ú
ú
ú
ú
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ê
ê
ê
ê
ë
é
-
-
--
---
=
==Atk
u
1
u
2 u
4
v
4v
2
v
1

28. 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
´
ú
ú
ú
ú
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ê
ê
ê
ê
ë
é
-
-
--
---
=
==Atk
u
3
u
4 u
2
v
2v
4
v
3

29. Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of
freedom
0
14433 ===== vvuvu
Notice that
Hence we need to calculate only a small (3x3) stiffness matrix
7
10
4.102.0
0983.045.0
2.045.0983.0
´
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
-
=K
u
1
u
2
v
2
u
1
u
2
v
2

30. Step 5: Compute consistent nodal loads
ï
þ
ï
ý
ü
ï
î
ï
í
ì
=
y
x
x
f
f
f
f
2
2
1
ï
þ
ï
ý
ü
ï
î
ï
í
ì
=
yf
2
0
0
y
Sy ff
2
1000
2 +-=
The consistent nodal load due to traction on the edge 3-2
lb
x
dx
x
dxN
tdxNf
x
x
x
S
y
225
2
9
50
2
50
3
150
)5.0)(300(
)300(
3
0
2
3
0
3
0
233
3
0
23
3
2
-=÷
ø
ö
ç
è
æ
-=
ú
û
ù
ê
ë
é
-=
-=
-=
-=
ò
ò
ò
=
=
-
=
-
3 2
3
23
2
x
N =
-

31. lb
ff
y
Sy
1225
1000
2
2
-=
+-=
Hence
Step 6: Solve the system equations to obtain the unknown nodal loads
fdK=
ï
þ
ï
ý
ü
ï
î
ï
í
ì
-
=
ï
þ
ï
ý
ü
ï
î
ï
í
ì
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
-
´
1225
0
0
4.102.0
0983.045.0
2.045.0983.0
10
2
2
1
7
v
u
u
Solve to get
ï
þ
ï
ý
ü
ï
î
ï
í
ì
´-
´
´
=
ï
þ
ï
ý
ü
ï
î
ï
í
ì
-
-
-
in
in
in
v
u
u
4
4
4
2
2
1
109084.0
101069.0
102337.0

32. Step 7: Compute the stresses in the elements
)1()1()1(
dBD=s
With
[ ]
[ ]00109084.0101069.00102337.0
d
444
442211
)1(
---
´-´´=
= vuvuvu
T
Calculate
psi
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
-
-
=
1.76
1.1391
1.114
)1(
s
In Element #1

33. )2()2()2(
dBD=s
With
[ ]
[ ]
44
224433
)2(
109084.0101069.00000
d
--
´-´=
= vuvuvu
T
Calculate
psi
ú
ú
ú
û
ù
ê
ê
ê
ë
é
-
=
35.363
52.28
1.114
)2(
s
In Element #2
Notice that the stresses are constant in each element