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

MANE 4240 & CIVL 4240
Introduction to Finite Elements
Prof. Suvranu De
Constant Strain
Triangle (CST)
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
Finite element formulation for 2D:
Step 1: Divide the body into finite elements connected to each
other through special points (“nodes”)
py
v3
3
px
4
3
u3
u 1 
v4
2
v 
v2
Element ‘e’
v
1 

1
4
u 2 
u
u4
ST
 
v1
2 u2
v 2 
y
d 
x
y
u3 

Su
u1
v 3 
1
 
v
x
x
u 4 
u
v 
 4
u (x,y)  N1(x,y) u1  N 2 (x,y) u 2  N 3(x,y) u 3  N 4 (x,y) u 4
v (x,y)  N1(x,y) v1  N 2 (x,y) v2  N 3(x,y) v3  N 4 (x,y) v4
u (x, y)  N 1
u

v (x, y)  0
uNd
0
N2
0
N3
0
N4
N1
0
N2
0
N3
0
u 1 
v 
 1
u 2 
 
0  v 2 
 

N 4  u 3 
v 3 
 
u 4 
v 
 4
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN
EACH ELEMENT
Approximation of the strain in element ‘e’
N 3 (x, y)
N 2 (x, y)
N 4 (x, y)
u (x, y) N1(x, y)
x 

u1 
u2 
u3 
u4
x
x
x
x
x
N 3 (x, y)
N 2 (x, y)
N 4 (x, y)
v (x, y) N1(x, y)
y 

v1 
v2 
v3 
v4
y
y
y
y
y
N1(x, y)
u (x, y) v (x, y) N1(x, y)
 xy 


u1 
v1  ......
y
x
y
x
x 
 
   y 
 xy 
 
u 1 
v 
 N 1(x, y)
 1 
N 3 (x, y)
N 2 (x, y)
N 4 (x, y)
0
0
0
0

 u 2 

x

x

x

x

 

N
(x,
y)

N
(x,
y)

N
(x,
y)

N
(x,
y)
3
1
2
4
 v 2 

0
0
0
0
 

y
y
y
y  u 3 
 N (x, y) N (x, y) N (x, y) N (x, y) N (x, y) N (x, y) N (x, y) N (x, y)
3
3
1
2
2
4
4
 1
 v 3 
x
y
x
y
x
y
x  u 
 y
  4 
B
v 
 4
ε Bd
Summary: For each element
Displacement approximation in terms of shape functions
uNd
Strain approximation in terms of strain-displacement matrix
ε Bd
Stress approximation
  DB d
Element stiffness matrix
k   e B D B dV
T
V
Element nodal load vector
f   e N X dV   e N T S dS
V
ST

 

 

T
f
T
b
f
S
Constant Strain Triangle (CST) : Simplest 2D finite element
v1
1
(x1,y1)
v2
y
u1
v3
(x3,y3)
v
(x,y)
2 (x2,y2)
u
3
u3
u2
x
• 3 nodes per element
• 2 dofs per node (each node can move in x- and y- directions)
• Hence 6 dofs per element
The displacement approximation in terms of shape functions is
u (x,y)  N1u1  N 2 u 2  N 3u 3
v(x,y)  N1v1  N 2 v 2  N 3v 3
u (x, y)  N1
u

v (x, y)  0
0
N1
N2
0
0
N2
N3
0
u 21  N 26 d 61
 N1
N
0
0
N2
0
N3
N1
0
N2
0
0
N 3 
u1 
v 
 1
0  u 2 
 
N 3  v 2 
u 3 
 
 v 3 
Formula for the shape functions are
v1
v3
1
u1
(x3,y3)
(x1,y1)
u3
v2 v
u
3
y
(x,y)
where
2 (x2,y2)
u2
x
a1  b1 x  c1 y
N1 
2A
a 2  b2 x  c2 y
N2 
2A
a3  b3 x  c3 y
N3 
2A
1 x 1
1
A  area of triangle  det 1 x 2
2
1 x 3
y1 
y 2 
y 3 
a1  x 2 y 3  x3 y 2
b1  y 2  y 3
c1  x3  x 2
a 2  x3 y1  x1 y 3
a3  x1 y 2  x 2 y1
b2  y 3  y1
b3  y1  y 2
c 2  x1  x3
c3  x 2  x1
Properties of the shape functions:
1. The shape functions N1, N2 and N3 are linear functions of x
and y
N2
N1
1
N3
1
1
1
1
3
3
y
1
2
3
2
x
1 at node' i'
Ni  
0 at other nodes
2
2. At every point in the domain
3
N
i 1
i
1
3
N x
i 1
i
i
3
N y
i 1
i
i
x
y
3. Geometric interpretation of the shape functions
At any point P(x,y) that the shape functions are evaluated,
P (x,y)
1
A3
y
A2
A1
2
x
3
A1
N1 
A
A2
N2 
A
A3
N3 
A
Approximation of the strains
 u 
 x


 x  

   v

  y 
 Bd

y
  

 xy   u v 
 y  x 


 N1(x,y)
0

 x
N1(x, y)
B 0

y
 N (x,y) N (x, y)
1
 1
x
 y
N 2 (x, y)
x
0
N 2 (x, y)
y
b1 0 b2 0 b3 0 
1
  0 c1 0 c2 0 c3 
2A
c1 b1 c2 b2 c3 b3 
0
N 2 (x, y)
y
N 2 (x, y)
x

0 

N 3 (x, y)
0
y 
N 3 (x, y) N 3 (x, y)

y
x 
N 3 (x, y)
x
Inside each element, all components of strain are constant: hence
the name Constant Strain Triangle
Element stresses (constant inside each element)
  DB d
IMPORTANT NOTE:
1. The displacement field is continuous across element
boundaries
2. The strains and stresses are NOT continuous across element
boundaries
Element stiffness matrix
t
k   e B D B dV
T
V
Since B is constant
A
k  B D B e dV  B D B At
T
T
V
t=thickness of the element
A=surface area of the element
Element nodal load vector
f   e N X dV   e N T S dS
V
ST

 

 

T
f
T
b
f
S
Element nodal load vector due to body forces
f b   e N X dV  t  e N X dA
T
T
V
A
fb1y
1
y
fb2y
fb1x
Xb
(x,y)
Xa
fb2x
2
x
fb3y
3
fb3x
t N X
1
a

Ae

f
 b1x 
 f  t  e N1 X b
 b1 y   A
 f b 2 x  t  e N 2 X a
fb 
 A
 f b 2 y  t Ae N 2 X b
 f b3 x  

 t Ae N 3 X a
 f b 3 y  
t Ae N 3 X b
dA

dA

dA

dA

dA

dA

EXAMPLE:
If Xa=1 and Xb=0
t N X
1
a

Ae

 f b1x 
 f   t  e N1 X b
 b1 y   A
 f b 2 x  t  e N 2 X a
fb 
 A
 f b 2 y  t Ae N 2 X b
 f b3 x  

 t Ae N 3 X a
 f b 3 y  
t Ae N 3 X b
dA
 tA 
 

t  e N 1 dA
3

A


dA
 
0
  0 
dA t N dA  tA 
   Ae 2    3 
0
dA 
 0
  t N dA   tA 
dA  Ae 3   
  3 
0
 
0
dA

Element nodal load vector due to traction
f S   e N T S dS
T
ST
EXAMPLE:
fS1y
f S  t
fS3y
1
l13
fS1x
3
y
2
x
fS3x
N
e
T
along 13
T S dS
Element nodal load vector due to traction
EXAMPLE:
f S  t
fS2y
l23
(2,2)
2
y
fS3y
1
(0,0)
fS2x
1
TS   
0
3 f
(2,0) S3x
T
x
N
e
f S2 x  t 
l 2 3
along 23
Similarly, compute
f S3 x  t
f S3 y  0
1
N 2 along 23 (1) dy
e
1
 t    2 1  t
 2
f S2 y  0
T S dS
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
y
3
1000 lb
300 psi
2
El 2
2 in
El 1
1
4
Thickness (t) = 0.5 in
E= 30×106 psi
n=0.25
x
3 in
(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


1
n
0
3.2 0.8 0 


E
0.8 3.2 0  107 psi


D
n
1
0


1 n 2 
1 n  
0 1.2
0 0
  0
2 

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
Recall
b1 0 b2 0 b3 0 
1
B   0 c1 0 c2 0 c3 
2A
c1 b1 c2 b2 c3 b3 
For Element #1:
2(2)
with
b1  y2  y3
c1  x3  x2
b2  y3  y1 b3  y1  y2
c2  x1  x3 c3  x2  x1
y1  0; y2  2; y3  0
x1  3; x2  3; x3  0
Hence
b1  2 b2  0 b3  2
c1  3 c2  3 c3  0
4(3)
1(1) Therefore
(local numbers within brackets)
For Element #2:
 2 0 0 0 2 0
1
( 2)
B   0 3 0  3 0 0
6
 3  2  3 0 0 2
 2 0 0 0 2 0 
1
(1)
B   0  3 0 3 0 0 
6
 3 2 3 0 0  2
Step 3: Compute element stiffness matrices
k  AtB
(1)
(1) T
D B  (3)(0.5)B
(1)
(1) T
DB
(1)
0.9833  0.5  0.45 0.2  0.5333

1.4
0.3  1.2
0.2


0.45
0
0

1.2
 0 .2


0.5333


u1
v1
u2
v2
u4
0.3 
 0.2
 0.3
7
 10
0 
0 

0.2 
v4
k
( 2)
 AtB
( 2) T
D B  (3)(0.5)B
( 2)
( 2) T
DB
( 2)
0.9833  0.5  0.45 0.2  0.5333

1.4
0.3  1.2
0.2


0.45
0
0

1.2
 0 .2


0.5333


u3
v3
u4
v4
u2
0.3 
 0.2
 0.3
7
 10
0 
0 

0.2 
v2
Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of
freedom
Notice that
u3  v3  u4  v4  v1  0
Hence we need to calculate only a small (3x3) stiffness matrix
u1
 0.983  0.45 0.2
K   0.45 0.983 0   107 u2
 0.2
0
1.4 
v2
u1
u2
v2
Step 5: Compute consistent nodal loads
 f1 x   0 
   
f   f2x    0 
f  f 
 2y   2y 
f 2 y  1000 f S2 y
The consistent nodal load due to traction on the edge 3-2
f S2 y  
3
x 0
N 3 3 2 (300)tdx
 (300)(0.5) 
3
x 0
x
3
N 3 3 2 dx
x
 150
dx
x 0 3
3
3
N 2 3 2 
 x2 
9
 50    50    225 lb
 2
 2 0
3
2
Hence
f 2 y  1000 f S2 y
 1225 lb
Step 6: Solve the system equations to obtain the unknown nodal loads
Kd  f
 0.983  0.45 0.2  u1   0 
  

107   0.45 0.983 0  u 2    0 
 

 0.2
0
1.4  
v2   1225
Solve to get
4
u1   0.233710 in 

  
4
u

0
.
1069

10
in
 2 

v   0.908410 4 in
 2 

Step 7: Compute the stresses in the elements
In Element #1
 (1)  D B(1) d(1)
With
d
(1) T
 u1 v1 u2 v2 u4 v4 

 0.2337104 0 0.1069104
Calculate
 (1)
  114.1 
  1391.1 psi
  76.1 
 0.9084104 0 0

In Element #2
 (2)  D B(2) d(2)
With
d
( 2)T
 u3 v3 u4 v4 u2 v2 

 0 0 0 0 0.1069104
 0.9084104
Calculate
 ( 2)
 114.1 
  28.52  psi
 363.35
Notice that the stresses are constant in each element
