Lst Elements
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Linear Strain Triangular Elements
10/14/08
Linear Strain Triangles
1
Degree of polynomial p 1
2
10/14/08
Number of terms n 3
6
Linear Strain Triangles
CST
LST
2
Degree of polynomial
10/14/08
p
Number of terms n
3
10
4
15
Linear Strain Triangles
QST
3
constant linear quadratic cubic
1 x x2 x x x
5
4
3
2
x y
x y
x y
xy 2
x y 3
y2
xy
3
4
y
x y
2
2
2
y xy
2
x y
3
3
3
y xy
4
4
y
5
quartic fifth order
Pascal’s Triangle 10/14/08
Linear Strain Triangles
4
v3
Basic 6 noded triangular element u3 3 v4
v5 5
u4 u5
4
v1
v2 u2
v6 u6 u1
2
6
1 10/14/08
Linear Strain Triangles
5
u1 d1 v1 d 2 u 2 d 3 v 2 { d} = = d 4 u 3 d 5 d 6 v 4 10/14/08
Linear Strain Triangles
6
u( x, y ) = a1 + a 2 x + a 3 y + a 4 x + a 5 xy + a 6 y 2
2
v( x, y ) = a 7 + a8 x + a 9 y + a10 x + a11xy + a12 y 2
10/14/08
Linear Strain Triangles
2
7
u 1 x y x 2 { ψ} = = v 0 0 0 0
10/14/08
xy y 2
0 0 0
0
1 x y x2
0
Linear Strain Triangles
0
a1 a2 a3 a4 a 5 0 0 a6 xy y 2 a 7 a8 a 9 a10 a11 a12 8
u 1 x y x 2 { ψ} = = v 0 0 0 0
xy y 2
0 0 0
0
1 x y x2
0
{ ψ } = [ M ]{ a} *
10/14/08
Linear Strain Triangles
0
a1 a 2 a3 a4 a5 0 0 a6 2 xy y a 7 a8 a9 a 10 a11 a 12 9
u 1 1 u 2 1 u 6 1 = v 1 0 v 2 0 v 0 6
10/14/08
x1 x2 x6 0 0 0
y1 y2 y6 0 0 0
x12 x 22 x 62 0 0 0
x1 y 1 x2y 2 x6y 6 0 0 0
y 12 x 22 y 62 0 0 0
0 0 0 1 1 1
0 0 0 x1 x2 x6
Linear Strain Triangles
0 0 0 y1 y2 y6
0 0 0 x12 x 22 x 62
0 0 0 x1 y 1 x2y 2 x6y 6
a1 a 2 0 a3 0 a4 a5 0 a6 2 y1 a7 2 y 2 a8 a9 2 y 6 a10 a11 a 12 10
a1 a 2 a 3 1 a 4 1 a5 a 6 1 = a 7 0 a 8 0 a9 a 0 10 a11 a 12 10/14/08
x1 x2 x6 0 0 0
y1 y2 y6 0 0 0
2 1 2 2
x x x 62 0 0 0
x1y 1 x2y 2 x6y 6 0 0 0
2 1 2 2
y 0 0 x 0 0 y 62 0 0 0 1 x1 0 1 x2 0 1 x6
Linear Strain Triangles
0 0 0 y1 y2 y6
0 0 0 x12 x 22 x 62
0 0 0 x1 y 1 x2y 2 x6y 6
0 0 0 y 12 2 y2 2 y 6
−1
u1 u 2 u 6 v1 v 2 v 6
11
{ ψ} = [ M ]{ a} *
{ a} = [ X] { d} −1
{ ψ} = [ N]{ d}
[ N ] = [ M ] [ X] *
10/14/08
−1
Linear Strain Triangles
12
Strain/Displacement ∂u ∂x εx ∂v { ε} = ε y = γ ∂ y xy ∂ u ∂ v + ∂ y ∂ x 10/14/08
Linear Strain Triangles
13
a1 0 1 0 2x y 0 0 0 0 0 0 0 a 2 { ε} = 0 0 0 0 0 0 0 0 1 0 x 2 y 0 0 1 0 x 2 y 0 1 0 2 x y 0 a 12
10/14/08
Linear Strain Triangles
14
{ ε} = [ M′] { a} { ε} = [ B] { d} { a} = [ X] { d} −1
[ B] = [ M′][ X] 10/14/08
−1
Linear Strain Triangles
15
σx εx { σ} = σ y = [ D] ε y τ γ xy xy
10/14/08
Linear Strain Triangles
16
[ k ] = ∫∫∫[ B] [ D] [ B] dV T
v
10/14/08
Linear Strain Triangles
17
β1 0 β 2 1 [ B] = 0 γ 1 0 2A γ 1 β1 γ 2
0
β3
0
β4
0
β5
0
β6
γ2
0
γ3
0
γ4
0
γ5
0
β2
γ3
β3
γ4
β4
γ5
β5
γ6
0 γ 6 β 6
β i = β i ( x, y , x c , y c ) γ i = γ i ( x, y , x c , y c )
10/14/08
Linear Strain Triangles
18
Example 9.2 ( 0, h )
h 0, 2
( 0,0 )
10/14/08
3
5
1
4 b , h 2 2
6 b ,0 2 Linear Strain Triangles
2
( b,0) 19
u( x, y ) = a1 + a 2 x + a 3 y + a 4 x + a 5 xy + a 6 y 2
2
2 2 ( ) v x, y = a 7 + a8 x + a 9 y + a10 x + a11xy + a12 y
10/14/08
Linear Strain Triangles
20
u1 = u( 0,0 ) = a1
u 2 = u( b,0 ) = a1 + a 2 b + a 4 b 2
u 3 = u( 0, h ) = a1 + a 3 h + a 6 h 2 2
b h b h b b h h u 4 = u , = a1 + a 2 + a 3 + a 4 + a 5 + a 6 2 2 2 2 2 2 2 2 h h h u 5 = u 0, = a1 + a 3 + a 6 2 2 2
2
b b b u 6 = u ,0 = a1 + a 2 + a4 2 2 2
2
10/14/08
Linear Strain Triangles
21
2
a1 = u1 4u 6 − 3u 4 − u 2 a2 = b 4u 5 − 3u1 − u 3 a3 = h 2( u 2 − 2u 6 + u1 ) a4 = b2 4( u1 + u4 − u5 − u6 ) a5 = bh 2( u 3 − 2u 5 + u1 ) a6 = h2 10/14/08
Linear Strain Triangles
22
a7 = v1 4v6 − 3v4 − v 2 a8 = b 4v5 − 3v1 − v 3 a9 = h 2( v 2 − 2 v 6 + v 1 ) a 10 = 2 b 4( v1 + v 4 − v 5 − v 6 ) a 11 = bh 2( v 3 − 2 v 5 + v 1 ) a 12 = 2 h 10/14/08
Linear Strain Triangles
23
4u 6 − 3u 4 − u 2 u = u1 + x+ b 4u 5 − 3u1 − u 3 2( u 2 − 2u 6 + u1 ) 2 y+ x + 2 h b 4( u1 + u4 − u5 − u6 ) 2( u 3 − 2u 5 + u1 ) 2 xy + y 2 bh h
10/14/08
Linear Strain Triangles
24
4v6 − 3v4 − v 2 v = v1 + x+ b 4v5 − 3v1 − v 3 2( v 2 − 2 v 6 + v 1 ) 2 y+ x + 2 h b 4( v1 + v 4 − v 5 − v 6 ) 2( v 3 − 2 v 5 + v 1 ) 2 xy + y 2 bh h
10/14/08
Linear Strain Triangles
25
u = v
N1 0
10/14/08
0
N2
0
N3
0
N4
0
N5
0
N6
N1
0
N2
0
N3
0
N4
0
N5
0
Linear Strain Triangles
u1 v1 0 u 2 N 6 v 6
26
2
y y x x x y N 1 = 1 − 3 − 3 + 2 + 4 + 2 b b b h h h x x N 2 = − + 2 b b
2
y y N 3 = − + 2 h h x y N 4 = 4 b h
2
y y x y N 5 = 4 − 4 − 4 b h h h
2
2
2
x x x y N 6 = 4 − 4 − 4 b b b h
10/14/08
Linear Strain Triangles
27
x1 = 0
y1 = 0
x2 = b
y2 = 0
x3 = 0
y3 = h
b x4 = 2
h y4 = 2 h y5 = 2
x5 = 0 b x6 = 2 10/14/08
y6 = 0
Linear Strain Triangles
28
2
y y x x x y N 1 = 1 − 3 − 3 + 2 + 4 + 2 b b b h h h
2
N1 ( 0,0 ) = 1
N1 ( b,0) = 0
N1 ( 0, h ) = 0 b h N1 , = 0 2 2 h N1 0, = 0 2 b N1 ,0 = 0 2
10/14/08
Linear Strain Triangles
29
x x N 2 = − + 2 b b
2
N 2 ( 0,0 ) = 0
N 2 ( b,0) = 1
N 2 ( 0, h ) = 0 b h N2 , = 0 2 2 h N 2 0, = 0 2 b N 2 ,0 = 0 2
10/14/08
Linear Strain Triangles
30
y y N 3 = − + 2 h h
2
N 3 ( 0,0 ) = 0
N 3 ( b,0) = 0
N 3 ( 0, h ) = 1 b h N3 , = 0 2 2 h N 3 0, = 0 2 b N 3 ,0 = 0 2 10/14/08
Linear Strain Triangles
31
x y N 4 = 4 b h N 4 ( 0,0 ) = 0
N 4 ( b,0) = 0
N 4 ( 0, h ) = 0 b h N4 , = 1 2 2 h N 4 0, = 0 2 b N 4 ,0 = 0 2
10/14/08
Linear Strain Triangles
32
y y x y N 5 = 4 + 4 − 4 b h h h
2
N 5 ( 0,0 ) = 0
N 5 ( b,0) = 0
N 5 ( 0, h ) = 0 b h N5 , = 0 2 2 h N 5 0, = 1 2 b N 5 ,0 = 0 2 10/14/08
Linear Strain Triangles
33
2
x x x y N 6 = 4 − 4 − 4 b b b h N 6 ( 0,0) = 0
N 6 ( b,0 ) = 0
N 6 ( 0, h ) = 0 b h N6 , = 0 2 2 h N 6 0, = 0 2 b N 6 ,0 = 1 2
10/14/08
Linear Strain Triangles
34
N1 + N 2 + N 3 + N4 + N5 + N6 = 2
y y x x x y 1 − 3 − 3 + 2 + 4 + 2 b b b h h h
2
2
2
y y x x x y − + 2 − + 2 + 4 b b b h h h 2
2
y y x y x x x y + 4 − 4 − 4 + 4 − 4 − 4 = b h b b b h h h y 1 + ( − 3 − 1 + 4) + ( − 3 − 1 + 4) + h 2
2
y x y + ( 2 + 2 − 4) + ( 4 + 4 − 4 − 4) b b h h
( 2 + 2 − 4 ) x
∑N
i
10/14/08
= N1 + N 2 + N 3 + N4 + N5 + N6 = 1 Linear Strain Triangles
35
β 1 1 [ B] = 0 2A γ 1
0
β2
0
β3
0
β4
0
β5
0
β6
γ1
0
γ2
0
γ3
0
γ4
0
γ5
0
β1
γ2
β2
γ3
β3
γ4
β4
γ5
β5
γ6
0 γ6 β 6
βi = βi (x, y ) γ i = γ i (x, y )
10/14/08
Linear Strain Triangles
36
β 1 1 [ B] = 0 2A γ 1
0
β2
0
β3
0
β4
0
β5
0
β6
γ1
0
γ2
0
γ3
0
γ4
0
γ5
0
β1
γ2
β2
γ3
β3
γ4
β4
γ5
β5
γ6
0 γ6 β 6
u = N1u1 + N 2u 2 + N 6 u6 ∂N 6 ∂N 2 ∂u ∂N 1 εx = = u1 + u2 + u6 ∂x ∂x ∂x ∂x εx =
10/14/08
1 ( β 1 u 1 + β 2 u 2 + β 6 u 6 ) 2A Linear Strain Triangles
37
βi ∂N i = 2A ∂x
10/14/08
Linear Strain Triangles
i = 1, 6
38
∂N i βi = 2A ∂x 2 A = bh
i = 1, 6
4 hx 3 4x 4 y β 1 = bh − + 2 + = − 3 h + + 4y bh b b b 4 hx 1 4x β 2 = bh − + 2 = − h + b b b β3 = 0 4y β4 = bh β 5 = −4 y 4 8x 4 y 8 hx β 6 = bh − 2 − − 4y = 4h − b b b bh 10/14/08
Linear Strain Triangles
39
∂N i γ i = 2A ∂y
i = 1, 6
4 by γ 1 = −3 b + 4 x + h γ2 = 0 4 by γ 3 = −b + h γ 4 = 4x 8 by γ 5 = 4b − 4x − h γ 6 = −4 x 10/14/08
Linear Strain Triangles
40
Comparison of CST and LST Elements
1 Linear Strain Triangle 6 Nodes 12 D-O-F 10/14/08
4 Constant Strain Triangles 6 Nodes 12 D-O-F
Linear Strain Triangles
41
4 x 16 mesh 48 in
12 in
Parabolic Load 40 kip (Total)
10/14/08
Linear Strain Triangles
42
Test Run
# of Nodes
4 x 16 Mesh
85
160
128 CST
8 x 32 Mesh
297
576
512 CST
2x8
85
160
32 LST
4 x 16
297
576
128 LST
10/14/08
# of D-O-F # of Elements
Linear Strain Triangles
43
Tip Max Deflection Stress x-locationy-location (in) (ksi)
Run
D-O-F
1
160
-0.29555 67.236
2.250
11.250
2
576
-0.33850 81.302
1.125
11.630
3
160
-0.33470 58.885
4.500
10.500
4
576
-0.35159 69.956
2.250
11.250
Exact Solution
10/14/08
-0.36133
80.00
Linear Strain Triangles
0
12
44
Natural or Intrinsic Coordinates 1D Elements
10/14/08
Linear Strain Triangles
45
L O
P
1
2 L2
x1
L1
x = ξ 1x1 + ξ 2x2 x2 = x1 + L
10/14/08
Linear Strain Triangles
46
Natural coordinates:
ξ 1 = L1 / L ξ 2 = L2 / L L1 + L2 = L
Not independent:
10/14/08
ξ1 + ξ2 = 1
Linear Strain Triangles
47
L 2
1
ξ 1 = L1 / L 1
ξ 2 = L2 / L 1
10/14/08
Linear Strain Triangles
48
x = ξ 1x1 + ξ 2x2 e.g. ξ 1 = ξ 2 = 1/2 x = (x1 + x2)/2
1 1 1 ξ 1 = x x 1 x 2 ξ 2 ξ 1 1 x 2 − 1 1 = ξ 2 L − x 1 1 x 10/14/08
Linear Strain Triangles
49
φ is an arbitrary variable. For example, it could be displacement u.
φ1 { φ} = [ N] φ 2 [ N] = [ ξ 1 ξ 2 ] N1 = ξ1 N2 = ξ2 10/14/08
Linear Strain Triangles
50
L O x1
10/14/08
3
1 L/2
Linear Strain Triangles
2 L/2
51
φ = a1 ξ 12 + a2 ξ 22 + a3 ξ 1 ξ 2 3
1
2
N1 = ξ 1 (2 ξ 1 -1) 3
1
1
1
10/14/08
2
N2 = ξ 2 (2 ξ 2 -1)
3 2
3
Linear Strain Triangles
2
N3 = 4 ξ 1 ξ 2
52
Write N1 directly: ξ1 3
1
2
(2 ξ 1 -1) N1 = ξ 1 (2 ξ 1 -1) ξ 1 = 1 at node 1 ξ 1 = 1/2 at node 2 ξ 1 = 0 at node 3
10/14/08
(2 ξ 1 -1) = 1 at node 1 (2 ξ 1 -1) = 0 at node 2 (2 ξ 1 -1) = -1 at node 3
Linear Strain Triangles
53
N1 = ξ 1 - N3/2 N2 = ξ 2 - N3/2 N1 + N2 + N3 = ξ 1 + ξ 2 =1
10/14/08
Linear Strain Triangles
54
φ=u u = ξ1u1 + ξ 2u 2 ∂u εx = ∂x ∂ u ∂ u ∂ ξ1 ∂ u ∂ ξ 2 = + ∂ x ∂ ξ1 ∂ x ∂ ξ 2 ∂ x ∂u = u1 ∂ ξ1 ∂u = u2 ∂ ξ2 10/14/08
Linear Strain Triangles
55
1 1 1 ξ 1 = x x 1 x 2 ξ 2 ξ 1 1 x 2 − 1 1 = ξ 2 L − x 1 1 x ∂ ξ1 ∂ x2 − x 1 = =− ∂x ∂x L L ∂ ξ2 ∂ x − x1 1 = = ∂x ∂x L L 1 1 ε x = u1 − + u 2 L L 10/14/08
Linear Strain Triangles
56
[ k ] = ∫ AE [ B] [ B] dL T
L
k! l! ξ ξ dL = L 1 2 ∫L ( 1 + k + l )! k
l
0! = 1
10/14/08
Linear Strain Triangles
57
Natural or Intrinsic Coordinates CST
10/14/08
Linear Strain Triangles
58
Natural Coordinates for a Triangular Element (Area Coordinates) 3
y
side 1
side 2
A1
A2 A3 1
2
side 3
x 10/14/08
Linear Strain Triangles
59
ξ i = A1/A ξ j = A2/A ξ m= A3/A ξi + ξj + ξj = 1 10/14/08
Linear Strain Triangles
60
Natural Coordinates for a Triangular Element (Area Coordinates)
ξ i = 0
ξ i = 0.2 5 ξ i = 0.5
ξm = 1 3 or m
ξ m = 0.25 2 or j
1
= ξj
5 0.
0
25 0.
75 0.
= ξj
= ξj
= ξj
Linear Strain Triangles
ξm = 0 = ξj
ξ i = 0.7 5 ξ i = 1 1 or i
10/14/08
ξ m = 0.75 ξ m = 0.5
61
1 1 x = x i y y i 1 [ A] = x i y i
1 xm y m
1 xj yj 1 xj yj
ξ 1 ξ 2 ξ 3
1 xm y m
ξ 1 1 −1 ξ 2 = [ A ] x ξ y 3 10/14/08
Linear Strain Triangles
62
[ A]
−1
αi 1 = α j 2A α m
βi βj βm
γi γ j γ m
2 A = ( x j − x 1 ) ( y m − y i ) + ( x m − x 1 )( y 2 − y i )
If i-j-m counterclockwise!
10/14/08
Linear Strain Triangles
63
α i = x j y m − y jx m βi = y j − y m
α j = xm y i − y m xi β j = ym − yi
αm = xi y j − y i x j βm = y i − y j
γ i = xm − x j
γ j = x i − xm
γ m = x j − xi
10/14/08
Linear Strain Triangles
64
Shape Functions ξ i = Ni ξ j = Nj ξ m = Nm φ = Ni φ i + Nj φ j + Nm φ m
10/14/08
Linear Strain Triangles
65
Derivatives by the chain rule.
∂ φ ∂ φ ∂ ξ1 ∂ φ ∂ ξ 2 ∂ φ ∂ ξ3 = + + ∂ x ∂ ξ1 ∂ x ∂ ξ 2 ∂ x ∂ ξ 3 ∂ x ∂ φ ∂ φ ∂ ξ1 ∂ φ ∂ ξ 2 ∂ φ ∂ ξ3 = + + ∂ y ∂ ξ1 ∂ y ∂ ξ 2 ∂ y ∂ ξ 3 ∂ y
10/14/08
Linear Strain Triangles
66
∂ ξ1 βi = ∂ x 2A
βj ∂ ξ2 = ∂x 2A
∂ ξ1 γi = ∂ y 2A
∂ ξ2 γi = ∂y 2A
10/14/08
Linear Strain Triangles
∂ ξ 3 βm = ∂x 2A ∂ ξ3 γi = ∂y 2A
67
∂φ = φi ∂ ξi ∂φ = φj ∂ ξj ∂φ = φm ∂ ξm 10/14/08
Linear Strain Triangles
68
ξ 1 1 −1 ξ 2 = [ A ] x ξ y 3 α i βi γ i 1 −1 [ A] = α j β j γ j 2A α m β m γ m 1 ( α i + βi x + γ i y) ξ1 = 2A ∂ ξ1 ∂ α i + βi x + γ i y βi = = ∂x ∂x 2A 2A
10/14/08
Linear Strain Triangles
69
l! n! p! ∫∫A ξ ξ ξ dA = 2A (2 + l + n + p)! l n p i j m
Centroidal Coordinates:
(
r s r s r s r s x y dA = C A x y + x y + x r +s 1 1 2 2 2y 2 ∫∫
)
A
r+s Cr+s 10/14/08
1 0
2 1/12
3 1/30
4 1/30
Linear Strain Triangles
5 2/105 70
Natural or Intrinsic Coordinates LST
10/14/08
Linear Strain Triangles
71
Natural Coordinates for a Triangular Element (Area Coordinates) 3
y
side 1
side 2
A1
A2 A3 1
2
side 3
x 10/14/08
Linear Strain Triangles
72
ξ i = A1/A ξ j = A2/A ξ m= A3/A ξi + ξj + ξj = 1 10/14/08
Linear Strain Triangles
73
Natural Coordinates for a Triangular Element (Area Coordinates)
ξ i = 0
ξ i = 0.2 5 ξ i = 0.5
ξm = 1 3 or m
ξ m = 0.25 2 or j
1
= ξj
5 0.
0
25 0.
75 0.
= ξj
= ξj
= ξj
Linear Strain Triangles
ξm = 0 = ξj
ξ i = 0.7 5 ξ i = 1 1 or i
10/14/08
ξ m = 0.75 ξ m = 0.5
74
Natural Coordinates for a Triangular Element (Area Coordinates)
Centroid: ξ i = ξ j= ξ m = 1/3 10/14/08
Linear Strain Triangles
75
Node 1 2 3 4 5 6
10/14/08
ξ1
ξ2
ξ3
1 0 0 0 1/2 1/2
0 1 0 1/2 0 1/2
0 0 1 1/2 1/2 0
Linear Strain Triangles
76
Shape Functions Ni
1
2
_________________________________________________
1 2 3 4 5 6
10/14/08
1 0 0 0 0 0
0 1 0 0 0 0
3
4
5
6
_________________________________________________
0 0 1 0 0 0
Linear Strain Triangles
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
77
Can we write shape functions directly? ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 1/2 at node 4 ξ 1 = 1/2 at node 5 ξ 1 = 0 at node 6
10/14/08
N1 = ξ 1 ( 2ξ 1-1 )
Linear Strain Triangles
78
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 0 at node 5 ξ 2 = 1/2 at node 6
10/14/08
N2 = ξ 2 ( 2ξ 2-1 )
Linear Strain Triangles
79
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 1/2 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 0 at node 6
10/14/08
N3 = ξ 3 ( 2ξ 3-1 )
Linear Strain Triangles
80
ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 0 at node 4 ξ 1 = 1/2 at node 5 ξ 1 = 1/2 at node 6
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 0 at node 5 ξ 2 = 1/2 at node 6
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 1/2 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 0 at node 6
N4 = 4 ξ 2 ξ 3 N5 = 4 ξ 1 ξ 3 N6 = 4 ξ 1 ξ 2
10/14/08
Linear Strain Triangles
81
ξ1 = s ξ2 = t ξ3 = 1 − s − t
10/14/08
Linear Strain Triangles
82
Step 1 -Select Element Type
x = a1 + a 2s + a 3 t + a4 s + a 5s t + a6 t 2
2
y = a 7 + a 8 s + a 9 t + a 10 s + a 11 s t + a 12 t 2
10/14/08
Linear Strain Triangles
2
83
Step 1 -Select Element Type
x N1 = y 0
10/14/08
0
N2
0
N3
0
N4
0
N5
0
N6
N1
0
N2
0
N3
0
N4
0
N5
0
Linear Strain Triangles
x1 y 1 x 2 0 y 2 N6 x6 y 6
84
Step 1 -Select Element Type ∂ N i ∂ N i ∂ ξ1 ∂ N i ∂ ξ 2 ∂ N i ∂ ξ 3 = + + ∂s ∂ ξ1 ∂ s ∂ ξ 2 ∂ s ∂ ξ 3 ∂ s ∂ Ni ∂ Ni ∂ Ni = − ∂s ∂ ξ1 ∂ ξ 3 ∂ Ni ∂ Ni ∂ Ni = − ∂t ∂ ξ2 ∂ ξ3 10/14/08
Linear Strain Triangles
85
Step 1 -Select Element Type N1 = ξ 1 ( 2ξ 1-1 ) N2 = ξ 2 ( 2ξ 2-1 ) N3 = ξ 3 ( 2ξ 3-1 ) N4 = 4 ξ 2 ξ 3 N5 = 4 ξ 1 ξ 3 N6 = 4 ξ 1 ξ 2
10/14/08
ξ1 = s ξ2 = t ξ3 = 1 − s − t
Linear Strain Triangles
86
Step 1 -Select Element Type N1 = s ( 2s - 1 ) N2 = t ( 2t - 1 ) N3 = (1 - s - t) ( 1 - 2s - 2t) N4 = 4 t (1 - s - t) N5 = 4 s (1 - s - t) N6 = 4 s t 10/14/08
Linear Strain Triangles
87
Step 2 -Select Displacement Function
u N1 = v 0
10/14/08
0
N2
0
N3
0
N4
0
N5
0
N6
N1
0
N2
0
N3
0
N4
0
N5
0
Linear Strain Triangles
u1 v 1 u 2 0 v 2 N6 u 6 v6
88
Step 3 -Define Stress/Strain and Strain/Displacement ∂u ε x ∂x ∂v { ε} = ε y = γ ∂y xy ∂u ∂v + ∂y ∂x { ε} = [ B]{ d} 10/14/08
Linear Strain Triangles
89
Chain Rule:
∂f ∂f ∂ x ∂f = + ∂s ∂ x ∂s ∂ y ∂f ∂f ∂ x ∂f = + ∂t ∂x ∂t ∂y 10/14/08
Linear Strain Triangles
∂y ∂s ∂y ∂t 90
Cramer’s Rule: ∂f ∂s ∂f ∂t
∂f = ∂x ∂x ∂s ∂x ∂t 10/14/08
∂y ∂s ∂y ∂t ∂y ∂s ∂y ∂t
∂x ∂s ∂x ∂t
∂f = ∂y ∂x ∂s ∂x ∂t
Linear Strain Triangles
∂f ∂s ∂f ∂t ∂y ∂s ∂y ∂t 91
Jacobian matrix:
∂x ∂s [ J] = ∂ x ∂ t 10/14/08
∂y ∂s ∂y ∂ t
Linear Strain Triangles
92
Strains in terms of an operator matrix:
∂( ) ε x ∂x { ε} = ε y = 0 γ xy ∂( ) ∂y 10/14/08
Linear Strain Triangles
0 ∂ ( ) u ∂y v ∂( ) ∂x 93
∂( ) 1 ∂y ∂ ( ) ∂y ∂ ( ) = − ∂x ∂s ∂t J ∂t ∂s ∂( ) 1 ∂x ∂ ( ) ∂ x ∂ ( ) = − ∂y ∂t ∂ s J ∂ s ∂t 10/14/08
Linear Strain Triangles
94
∂y ∂ ( ) ∂y ∂ ( ) − ∂t ∂s ∂s ∂t εx 1 0 εy = γ J ∂x ∂ ( ) ∂x ∂ ( ) xy − ∂s ∂t ∂t ∂ s
10/14/08
0 ∂x ∂ ( ) ∂x ∂ ( ) u − ∂s ∂t ∂t ∂s v ∂y ∂ ( ) ∂y ∂ ( ) − ∂t ∂s ∂s ∂t
Linear Strain Triangles
95
{ ε} = [ D ′] [ N ] { d} ∂y ∂t 1 [ D′] = J ∂x ∂s 10/14/08
∂ ( ) ∂y ∂ ( ) − ∂s ∂s ∂t 0 ∂ ( ) ∂x ∂ ( ) − ∂t ∂t ∂s
0 ∂x ∂s ∂y ∂t
Linear Strain Triangles
∂ ( ) ∂x − ∂t ∂t ∂ ( ) ∂y − ∂s ∂s
∂( ) ∂s ∂( ) ∂t 96
Step 4 -Derive Element Stiffness Matrix and Equations
[ k ] = ∫∫ [ B ] [ D ] [ B ] T
t dx dy
A
f ( x , y ) dx dy = ∫∫ A
f ( s , t ) J ds dt ∫∫ A
[ k ] = ∫∫ [ B ] [ D ] [ B ] T
t J ds dt
A
10/14/08
Linear Strain Triangles
97
{ fb } = ∫ ∫ [ N] { Xb } t J ds dt T
A
{ fs } = ∫L [ N] { T} t J dL T
10/14/08
Linear Strain Triangles
98
Need to use numerical integration to evaluate the element stiffness matrix.
∫∫f (ξ , ξ 1
A
10/14/08
n
2
(
, ξ 3 ) dA = ∑ Wi f ξ1i , ξ 2i , ξ 2i i =1
Linear Strain Triangles
)
99
1-point Gauss Formula
Wi = 1 ξ 1i = 1/3 ξ 2i = 1/3 ξ 3i = 1/3 10/14/08
Linear Strain Triangles
100
3-point Gauss Formula
Wi = 1/3 ξ 1i = 1/3, 1/6, 1/6 ξ 2i = 1/6, 1/3 , 1/6 ξ 3i = 1/6 , 1/6, 1/3 10/14/08
Linear Strain Triangles
101
3-point Gauss Formula
Wi = 1/3 ξ 1i = 1/2, 1/2, 0 ξ 2i = 1/2, 0, 1/2 ξ 3i = 0, 1/2, 1/2 10/14/08
Linear Strain Triangles
102
4-point Gauss Formula Wi = -0.56250 ξ 1i = 1/3 ξ 2i = 1/3 ξ 3i = 1/3 Wi = 0.52083 33333 33333 ξ 1i = 0.6, 0.2, 0.2 ξ 2i = 0.2, 0.6, 0.2 ξ 3i = 0.2, 0.2, 0.6 10/14/08
Linear Strain Triangles
103
6-point Gauss Formula Wi = 0.10995 17436 55322 ξ 1i = 0.81684 75729 80459 ξ 2i = 0.09157 62135 09771 ξ 3i = 0.09157 62135 09771 Wi = 0.22338 15896 78011 ξ 1i = 0.10810 30181 68070 ξ 2i = 0.44594 84909 15965 ξ 3i = 0.44594 84909 15965 10/14/08
Linear Strain Triangles
104
7-point Gauss Formula Wi = 0.12593 91805 44827 ξ 1i = 0.79742 69853 53087 ξ 2i = 0.10128 65073 23456 ξ 3i = 0.10128 65073 23456 Wi = 0.13239 41527 88506 ξ 1i = 0.47014 20641 05115 ξ 2i = 0.47014 20641 05115 ξ 3i = 0.05971 58717 89770 10/14/08
Linear Strain Triangles
Wi = 0.225 ξ 1i = 1/3 ξ 2i = 1/3 ξ 3i = 1/3 105
Pascal’s Triangle:
Degree of polynomial 0
1 x x x x x5 10/14/08
x4 y
y
x y 3
2
2
xy
2
x y
4
y xy
2
3
1
x y 2
x3 y2
2
xy
3
3
x2 y3
3
y
2
4
y
x y4
4
y5
Linear Strain Triangles
5
106
Degree of polynomial Number of terms n p
10/14/08
1
3
2
6
3
10
4
15
Linear Strain Triangles
107
n
φ = ∑ a i ξ 1q ξ r2 ξ s3 i =1
q+r+s =p Quadratic Triangle n=6 p=2 ( p + 1)( p + 2 ) n= 2 φ = a 1 ξ 12 + a 2 ξ 22 + a 3 ξ 23 + a 4 ξ 1 ξ 2 + a 5 ξ 2 ξ 3 + a 6 ξ 3 ξ 1 Equivalent to : φ = b 1 + b 2 x + b 3 y + b 4 x 2 + d 5 xy + b 6 y 2 10/14/08
Linear Strain Triangles
108
Nodal Numbering Schemes: 3
3 6
5
2 1
1
3 8
4
2
7
9 10
6 2
10/14/08
1
4Linear Strain 5Triangles
109
Quadratic Triangle φ = a 1 ξ 12 + a 2 ξ 22 + a 3 ξ 23 + a 4 ξ 1 ξ 2 + a 5 ξ 2 ξ 3 + a 6 ξ 3 ξ 1
Node 1 2 3 4 5 6 10/14/08
ξ1
ξ2
ξ3
1 0 0 1/2 0 1/2
0 1 0 1/2 1/2 0
0 0 1 0 1/2 1/2
Linear Strain Triangles
110
Shape Functions
Ni
1
2
_________________________________________________
1 2 3 4 5 6
10/14/08
1 0 0 0 0 0
0 1 0 0 0 0
3
4
5
6
_________________________________________________
0 0 1 0 0 0
Linear Strain Triangles
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
111
Can we write shape functions directly? ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 1/2 at node 4 ξ 1 = 0 at node 5 ξ 1 = 1/2 at node 6
10/14/08
N1 = ξ 1 ( 2ξ 1-1 )
Linear Strain Triangles
112
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 1/2 at node 5 ξ 2 = 0 at node 6
10/14/08
N2 = ξ 2 ( 2ξ 2-1 )
Linear Strain Triangles
113
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 1 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 1/2 at node 6
10/14/08
N3 = ξ 3 ( 2ξ 3-1 )
Linear Strain Triangles
114
ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 1/2 at node 4 ξ 1 = 0 at node 5 ξ 1 = 1/2 at node 6
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 1/2 at node 5 ξ 2 = 0 at node 6
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 0 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 1/2 at node 6
N4 = 4 ξ 1 ξ 2 N5 = 4 ξ 2 ξ 3 N6 = 4 ξ 3 ξ 1
10/14/08
Linear Strain Triangles
115
∂u ∂N i εX = =∑ ui ∂x i =1 ∂x 3 ∂N i ∂N i ∂ξ j =∑ ∂x i = 1 ∂ ξ j ∂x 6
10/14/08
Linear Strain Triangles
116
Cubic Element
ξ i = 1, 0, 1/3, 2/3
N1 = 1/2 ξ 1(3 ξ 1 -1)(3 ξ 1 -2) N2 = 1/2 ξ 2(3 ξ 2 -1)(3 ξ 2 -2) N3 = 1/2 ξ 2(3 ξ 2 -1)(3 ξ 2 -2) N4 = 9/2 ξ 1 ξ 2(3 ξ 1-1) N5 = 9/2 ξ 2 ξ 1(3 ξ 2-1) N6 = 9/2 ξ 2 ξ 1(3 ξ 1-1) N7 = 9/2 ξ 2 ξ 3(3 ξ 2-1) N8 = 9/2 ξ 2 ξ 3(3 ξ 3-1) N9 = 9/2 ξ 1 ξ 3(3 ξ 1-1) N10 = 27 ξ 1 ξ 2 ξ 3 10/14/08
Linear Strain Triangles
117
10/14/08
Linear Strain Triangles
1
Degree of polynomial p 1
2
10/14/08
Number of terms n 3
6
Linear Strain Triangles
CST
LST
2
Degree of polynomial
10/14/08
p
Number of terms n
3
10
4
15
Linear Strain Triangles
QST
3
constant linear quadratic cubic
1 x x2 x x x
5
4
3
2
x y
x y
x y
xy 2
x y 3
y2
xy
3
4
y
x y
2
2
2
y xy
2
x y
3
3
3
y xy
4
4
y
5
quartic fifth order
Pascal’s Triangle 10/14/08
Linear Strain Triangles
4
v3
Basic 6 noded triangular element u3 3 v4
v5 5
u4 u5
4
v1
v2 u2
v6 u6 u1
2
6
1 10/14/08
Linear Strain Triangles
5
u1 d1 v1 d 2 u 2 d 3 v 2 { d} = = d 4 u 3 d 5 d 6 v 4 10/14/08
Linear Strain Triangles
6
u( x, y ) = a1 + a 2 x + a 3 y + a 4 x + a 5 xy + a 6 y 2
2
v( x, y ) = a 7 + a8 x + a 9 y + a10 x + a11xy + a12 y 2
10/14/08
Linear Strain Triangles
2
7
u 1 x y x 2 { ψ} = = v 0 0 0 0
10/14/08
xy y 2
0 0 0
0
1 x y x2
0
Linear Strain Triangles
0
a1 a2 a3 a4 a 5 0 0 a6 xy y 2 a 7 a8 a 9 a10 a11 a12 8
u 1 x y x 2 { ψ} = = v 0 0 0 0
xy y 2
0 0 0
0
1 x y x2
0
{ ψ } = [ M ]{ a} *
10/14/08
Linear Strain Triangles
0
a1 a 2 a3 a4 a5 0 0 a6 2 xy y a 7 a8 a9 a 10 a11 a 12 9
u 1 1 u 2 1 u 6 1 = v 1 0 v 2 0 v 0 6
10/14/08
x1 x2 x6 0 0 0
y1 y2 y6 0 0 0
x12 x 22 x 62 0 0 0
x1 y 1 x2y 2 x6y 6 0 0 0
y 12 x 22 y 62 0 0 0
0 0 0 1 1 1
0 0 0 x1 x2 x6
Linear Strain Triangles
0 0 0 y1 y2 y6
0 0 0 x12 x 22 x 62
0 0 0 x1 y 1 x2y 2 x6y 6
a1 a 2 0 a3 0 a4 a5 0 a6 2 y1 a7 2 y 2 a8 a9 2 y 6 a10 a11 a 12 10
a1 a 2 a 3 1 a 4 1 a5 a 6 1 = a 7 0 a 8 0 a9 a 0 10 a11 a 12 10/14/08
x1 x2 x6 0 0 0
y1 y2 y6 0 0 0
2 1 2 2
x x x 62 0 0 0
x1y 1 x2y 2 x6y 6 0 0 0
2 1 2 2
y 0 0 x 0 0 y 62 0 0 0 1 x1 0 1 x2 0 1 x6
Linear Strain Triangles
0 0 0 y1 y2 y6
0 0 0 x12 x 22 x 62
0 0 0 x1 y 1 x2y 2 x6y 6
0 0 0 y 12 2 y2 2 y 6
−1
u1 u 2 u 6 v1 v 2 v 6
11
{ ψ} = [ M ]{ a} *
{ a} = [ X] { d} −1
{ ψ} = [ N]{ d}
[ N ] = [ M ] [ X] *
10/14/08
−1
Linear Strain Triangles
12
Strain/Displacement ∂u ∂x εx ∂v { ε} = ε y = γ ∂ y xy ∂ u ∂ v + ∂ y ∂ x 10/14/08
Linear Strain Triangles
13
a1 0 1 0 2x y 0 0 0 0 0 0 0 a 2 { ε} = 0 0 0 0 0 0 0 0 1 0 x 2 y 0 0 1 0 x 2 y 0 1 0 2 x y 0 a 12
10/14/08
Linear Strain Triangles
14
{ ε} = [ M′] { a} { ε} = [ B] { d} { a} = [ X] { d} −1
[ B] = [ M′][ X] 10/14/08
−1
Linear Strain Triangles
15
σx εx { σ} = σ y = [ D] ε y τ γ xy xy
10/14/08
Linear Strain Triangles
16
[ k ] = ∫∫∫[ B] [ D] [ B] dV T
v
10/14/08
Linear Strain Triangles
17
β1 0 β 2 1 [ B] = 0 γ 1 0 2A γ 1 β1 γ 2
0
β3
0
β4
0
β5
0
β6
γ2
0
γ3
0
γ4
0
γ5
0
β2
γ3
β3
γ4
β4
γ5
β5
γ6
0 γ 6 β 6
β i = β i ( x, y , x c , y c ) γ i = γ i ( x, y , x c , y c )
10/14/08
Linear Strain Triangles
18
Example 9.2 ( 0, h )
h 0, 2
( 0,0 )
10/14/08
3
5
1
4 b , h 2 2
6 b ,0 2 Linear Strain Triangles
2
( b,0) 19
u( x, y ) = a1 + a 2 x + a 3 y + a 4 x + a 5 xy + a 6 y 2
2
2 2 ( ) v x, y = a 7 + a8 x + a 9 y + a10 x + a11xy + a12 y
10/14/08
Linear Strain Triangles
20
u1 = u( 0,0 ) = a1
u 2 = u( b,0 ) = a1 + a 2 b + a 4 b 2
u 3 = u( 0, h ) = a1 + a 3 h + a 6 h 2 2
b h b h b b h h u 4 = u , = a1 + a 2 + a 3 + a 4 + a 5 + a 6 2 2 2 2 2 2 2 2 h h h u 5 = u 0, = a1 + a 3 + a 6 2 2 2
2
b b b u 6 = u ,0 = a1 + a 2 + a4 2 2 2
2
10/14/08
Linear Strain Triangles
21
2
a1 = u1 4u 6 − 3u 4 − u 2 a2 = b 4u 5 − 3u1 − u 3 a3 = h 2( u 2 − 2u 6 + u1 ) a4 = b2 4( u1 + u4 − u5 − u6 ) a5 = bh 2( u 3 − 2u 5 + u1 ) a6 = h2 10/14/08
Linear Strain Triangles
22
a7 = v1 4v6 − 3v4 − v 2 a8 = b 4v5 − 3v1 − v 3 a9 = h 2( v 2 − 2 v 6 + v 1 ) a 10 = 2 b 4( v1 + v 4 − v 5 − v 6 ) a 11 = bh 2( v 3 − 2 v 5 + v 1 ) a 12 = 2 h 10/14/08
Linear Strain Triangles
23
4u 6 − 3u 4 − u 2 u = u1 + x+ b 4u 5 − 3u1 − u 3 2( u 2 − 2u 6 + u1 ) 2 y+ x + 2 h b 4( u1 + u4 − u5 − u6 ) 2( u 3 − 2u 5 + u1 ) 2 xy + y 2 bh h
10/14/08
Linear Strain Triangles
24
4v6 − 3v4 − v 2 v = v1 + x+ b 4v5 − 3v1 − v 3 2( v 2 − 2 v 6 + v 1 ) 2 y+ x + 2 h b 4( v1 + v 4 − v 5 − v 6 ) 2( v 3 − 2 v 5 + v 1 ) 2 xy + y 2 bh h
10/14/08
Linear Strain Triangles
25
u = v
N1 0
10/14/08
0
N2
0
N3
0
N4
0
N5
0
N6
N1
0
N2
0
N3
0
N4
0
N5
0
Linear Strain Triangles
u1 v1 0 u 2 N 6 v 6
26
2
y y x x x y N 1 = 1 − 3 − 3 + 2 + 4 + 2 b b b h h h x x N 2 = − + 2 b b
2
y y N 3 = − + 2 h h x y N 4 = 4 b h
2
y y x y N 5 = 4 − 4 − 4 b h h h
2
2
2
x x x y N 6 = 4 − 4 − 4 b b b h
10/14/08
Linear Strain Triangles
27
x1 = 0
y1 = 0
x2 = b
y2 = 0
x3 = 0
y3 = h
b x4 = 2
h y4 = 2 h y5 = 2
x5 = 0 b x6 = 2 10/14/08
y6 = 0
Linear Strain Triangles
28
2
y y x x x y N 1 = 1 − 3 − 3 + 2 + 4 + 2 b b b h h h
2
N1 ( 0,0 ) = 1
N1 ( b,0) = 0
N1 ( 0, h ) = 0 b h N1 , = 0 2 2 h N1 0, = 0 2 b N1 ,0 = 0 2
10/14/08
Linear Strain Triangles
29
x x N 2 = − + 2 b b
2
N 2 ( 0,0 ) = 0
N 2 ( b,0) = 1
N 2 ( 0, h ) = 0 b h N2 , = 0 2 2 h N 2 0, = 0 2 b N 2 ,0 = 0 2
10/14/08
Linear Strain Triangles
30
y y N 3 = − + 2 h h
2
N 3 ( 0,0 ) = 0
N 3 ( b,0) = 0
N 3 ( 0, h ) = 1 b h N3 , = 0 2 2 h N 3 0, = 0 2 b N 3 ,0 = 0 2 10/14/08
Linear Strain Triangles
31
x y N 4 = 4 b h N 4 ( 0,0 ) = 0
N 4 ( b,0) = 0
N 4 ( 0, h ) = 0 b h N4 , = 1 2 2 h N 4 0, = 0 2 b N 4 ,0 = 0 2
10/14/08
Linear Strain Triangles
32
y y x y N 5 = 4 + 4 − 4 b h h h
2
N 5 ( 0,0 ) = 0
N 5 ( b,0) = 0
N 5 ( 0, h ) = 0 b h N5 , = 0 2 2 h N 5 0, = 1 2 b N 5 ,0 = 0 2 10/14/08
Linear Strain Triangles
33
2
x x x y N 6 = 4 − 4 − 4 b b b h N 6 ( 0,0) = 0
N 6 ( b,0 ) = 0
N 6 ( 0, h ) = 0 b h N6 , = 0 2 2 h N 6 0, = 0 2 b N 6 ,0 = 1 2
10/14/08
Linear Strain Triangles
34
N1 + N 2 + N 3 + N4 + N5 + N6 = 2
y y x x x y 1 − 3 − 3 + 2 + 4 + 2 b b b h h h
2
2
2
y y x x x y − + 2 − + 2 + 4 b b b h h h 2
2
y y x y x x x y + 4 − 4 − 4 + 4 − 4 − 4 = b h b b b h h h y 1 + ( − 3 − 1 + 4) + ( − 3 − 1 + 4) + h 2
2
y x y + ( 2 + 2 − 4) + ( 4 + 4 − 4 − 4) b b h h
( 2 + 2 − 4 ) x
∑N
i
10/14/08
= N1 + N 2 + N 3 + N4 + N5 + N6 = 1 Linear Strain Triangles
35
β 1 1 [ B] = 0 2A γ 1
0
β2
0
β3
0
β4
0
β5
0
β6
γ1
0
γ2
0
γ3
0
γ4
0
γ5
0
β1
γ2
β2
γ3
β3
γ4
β4
γ5
β5
γ6
0 γ6 β 6
βi = βi (x, y ) γ i = γ i (x, y )
10/14/08
Linear Strain Triangles
36
β 1 1 [ B] = 0 2A γ 1
0
β2
0
β3
0
β4
0
β5
0
β6
γ1
0
γ2
0
γ3
0
γ4
0
γ5
0
β1
γ2
β2
γ3
β3
γ4
β4
γ5
β5
γ6
0 γ6 β 6
u = N1u1 + N 2u 2 + N 6 u6 ∂N 6 ∂N 2 ∂u ∂N 1 εx = = u1 + u2 + u6 ∂x ∂x ∂x ∂x εx =
10/14/08
1 ( β 1 u 1 + β 2 u 2 + β 6 u 6 ) 2A Linear Strain Triangles
37
βi ∂N i = 2A ∂x
10/14/08
Linear Strain Triangles
i = 1, 6
38
∂N i βi = 2A ∂x 2 A = bh
i = 1, 6
4 hx 3 4x 4 y β 1 = bh − + 2 + = − 3 h + + 4y bh b b b 4 hx 1 4x β 2 = bh − + 2 = − h + b b b β3 = 0 4y β4 = bh β 5 = −4 y 4 8x 4 y 8 hx β 6 = bh − 2 − − 4y = 4h − b b b bh 10/14/08
Linear Strain Triangles
39
∂N i γ i = 2A ∂y
i = 1, 6
4 by γ 1 = −3 b + 4 x + h γ2 = 0 4 by γ 3 = −b + h γ 4 = 4x 8 by γ 5 = 4b − 4x − h γ 6 = −4 x 10/14/08
Linear Strain Triangles
40
Comparison of CST and LST Elements
1 Linear Strain Triangle 6 Nodes 12 D-O-F 10/14/08
4 Constant Strain Triangles 6 Nodes 12 D-O-F
Linear Strain Triangles
41
4 x 16 mesh 48 in
12 in
Parabolic Load 40 kip (Total)
10/14/08
Linear Strain Triangles
42
Test Run
# of Nodes
4 x 16 Mesh
85
160
128 CST
8 x 32 Mesh
297
576
512 CST
2x8
85
160
32 LST
4 x 16
297
576
128 LST
10/14/08
# of D-O-F # of Elements
Linear Strain Triangles
43
Tip Max Deflection Stress x-locationy-location (in) (ksi)
Run
D-O-F
1
160
-0.29555 67.236
2.250
11.250
2
576
-0.33850 81.302
1.125
11.630
3
160
-0.33470 58.885
4.500
10.500
4
576
-0.35159 69.956
2.250
11.250
Exact Solution
10/14/08
-0.36133
80.00
Linear Strain Triangles
0
12
44
Natural or Intrinsic Coordinates 1D Elements
10/14/08
Linear Strain Triangles
45
L O
P
1
2 L2
x1
L1
x = ξ 1x1 + ξ 2x2 x2 = x1 + L
10/14/08
Linear Strain Triangles
46
Natural coordinates:
ξ 1 = L1 / L ξ 2 = L2 / L L1 + L2 = L
Not independent:
10/14/08
ξ1 + ξ2 = 1
Linear Strain Triangles
47
L 2
1
ξ 1 = L1 / L 1
ξ 2 = L2 / L 1
10/14/08
Linear Strain Triangles
48
x = ξ 1x1 + ξ 2x2 e.g. ξ 1 = ξ 2 = 1/2 x = (x1 + x2)/2
1 1 1 ξ 1 = x x 1 x 2 ξ 2 ξ 1 1 x 2 − 1 1 = ξ 2 L − x 1 1 x 10/14/08
Linear Strain Triangles
49
φ is an arbitrary variable. For example, it could be displacement u.
φ1 { φ} = [ N] φ 2 [ N] = [ ξ 1 ξ 2 ] N1 = ξ1 N2 = ξ2 10/14/08
Linear Strain Triangles
50
L O x1
10/14/08
3
1 L/2
Linear Strain Triangles
2 L/2
51
φ = a1 ξ 12 + a2 ξ 22 + a3 ξ 1 ξ 2 3
1
2
N1 = ξ 1 (2 ξ 1 -1) 3
1
1
1
10/14/08
2
N2 = ξ 2 (2 ξ 2 -1)
3 2
3
Linear Strain Triangles
2
N3 = 4 ξ 1 ξ 2
52
Write N1 directly: ξ1 3
1
2
(2 ξ 1 -1) N1 = ξ 1 (2 ξ 1 -1) ξ 1 = 1 at node 1 ξ 1 = 1/2 at node 2 ξ 1 = 0 at node 3
10/14/08
(2 ξ 1 -1) = 1 at node 1 (2 ξ 1 -1) = 0 at node 2 (2 ξ 1 -1) = -1 at node 3
Linear Strain Triangles
53
N1 = ξ 1 - N3/2 N2 = ξ 2 - N3/2 N1 + N2 + N3 = ξ 1 + ξ 2 =1
10/14/08
Linear Strain Triangles
54
φ=u u = ξ1u1 + ξ 2u 2 ∂u εx = ∂x ∂ u ∂ u ∂ ξ1 ∂ u ∂ ξ 2 = + ∂ x ∂ ξ1 ∂ x ∂ ξ 2 ∂ x ∂u = u1 ∂ ξ1 ∂u = u2 ∂ ξ2 10/14/08
Linear Strain Triangles
55
1 1 1 ξ 1 = x x 1 x 2 ξ 2 ξ 1 1 x 2 − 1 1 = ξ 2 L − x 1 1 x ∂ ξ1 ∂ x2 − x 1 = =− ∂x ∂x L L ∂ ξ2 ∂ x − x1 1 = = ∂x ∂x L L 1 1 ε x = u1 − + u 2 L L 10/14/08
Linear Strain Triangles
56
[ k ] = ∫ AE [ B] [ B] dL T
L
k! l! ξ ξ dL = L 1 2 ∫L ( 1 + k + l )! k
l
0! = 1
10/14/08
Linear Strain Triangles
57
Natural or Intrinsic Coordinates CST
10/14/08
Linear Strain Triangles
58
Natural Coordinates for a Triangular Element (Area Coordinates) 3
y
side 1
side 2
A1
A2 A3 1
2
side 3
x 10/14/08
Linear Strain Triangles
59
ξ i = A1/A ξ j = A2/A ξ m= A3/A ξi + ξj + ξj = 1 10/14/08
Linear Strain Triangles
60
Natural Coordinates for a Triangular Element (Area Coordinates)
ξ i = 0
ξ i = 0.2 5 ξ i = 0.5
ξm = 1 3 or m
ξ m = 0.25 2 or j
1
= ξj
5 0.
0
25 0.
75 0.
= ξj
= ξj
= ξj
Linear Strain Triangles
ξm = 0 = ξj
ξ i = 0.7 5 ξ i = 1 1 or i
10/14/08
ξ m = 0.75 ξ m = 0.5
61
1 1 x = x i y y i 1 [ A] = x i y i
1 xm y m
1 xj yj 1 xj yj
ξ 1 ξ 2 ξ 3
1 xm y m
ξ 1 1 −1 ξ 2 = [ A ] x ξ y 3 10/14/08
Linear Strain Triangles
62
[ A]
−1
αi 1 = α j 2A α m
βi βj βm
γi γ j γ m
2 A = ( x j − x 1 ) ( y m − y i ) + ( x m − x 1 )( y 2 − y i )
If i-j-m counterclockwise!
10/14/08
Linear Strain Triangles
63
α i = x j y m − y jx m βi = y j − y m
α j = xm y i − y m xi β j = ym − yi
αm = xi y j − y i x j βm = y i − y j
γ i = xm − x j
γ j = x i − xm
γ m = x j − xi
10/14/08
Linear Strain Triangles
64
Shape Functions ξ i = Ni ξ j = Nj ξ m = Nm φ = Ni φ i + Nj φ j + Nm φ m
10/14/08
Linear Strain Triangles
65
Derivatives by the chain rule.
∂ φ ∂ φ ∂ ξ1 ∂ φ ∂ ξ 2 ∂ φ ∂ ξ3 = + + ∂ x ∂ ξ1 ∂ x ∂ ξ 2 ∂ x ∂ ξ 3 ∂ x ∂ φ ∂ φ ∂ ξ1 ∂ φ ∂ ξ 2 ∂ φ ∂ ξ3 = + + ∂ y ∂ ξ1 ∂ y ∂ ξ 2 ∂ y ∂ ξ 3 ∂ y
10/14/08
Linear Strain Triangles
66
∂ ξ1 βi = ∂ x 2A
βj ∂ ξ2 = ∂x 2A
∂ ξ1 γi = ∂ y 2A
∂ ξ2 γi = ∂y 2A
10/14/08
Linear Strain Triangles
∂ ξ 3 βm = ∂x 2A ∂ ξ3 γi = ∂y 2A
67
∂φ = φi ∂ ξi ∂φ = φj ∂ ξj ∂φ = φm ∂ ξm 10/14/08
Linear Strain Triangles
68
ξ 1 1 −1 ξ 2 = [ A ] x ξ y 3 α i βi γ i 1 −1 [ A] = α j β j γ j 2A α m β m γ m 1 ( α i + βi x + γ i y) ξ1 = 2A ∂ ξ1 ∂ α i + βi x + γ i y βi = = ∂x ∂x 2A 2A
10/14/08
Linear Strain Triangles
69
l! n! p! ∫∫A ξ ξ ξ dA = 2A (2 + l + n + p)! l n p i j m
Centroidal Coordinates:
(
r s r s r s r s x y dA = C A x y + x y + x r +s 1 1 2 2 2y 2 ∫∫
)
A
r+s Cr+s 10/14/08
1 0
2 1/12
3 1/30
4 1/30
Linear Strain Triangles
5 2/105 70
Natural or Intrinsic Coordinates LST
10/14/08
Linear Strain Triangles
71
Natural Coordinates for a Triangular Element (Area Coordinates) 3
y
side 1
side 2
A1
A2 A3 1
2
side 3
x 10/14/08
Linear Strain Triangles
72
ξ i = A1/A ξ j = A2/A ξ m= A3/A ξi + ξj + ξj = 1 10/14/08
Linear Strain Triangles
73
Natural Coordinates for a Triangular Element (Area Coordinates)
ξ i = 0
ξ i = 0.2 5 ξ i = 0.5
ξm = 1 3 or m
ξ m = 0.25 2 or j
1
= ξj
5 0.
0
25 0.
75 0.
= ξj
= ξj
= ξj
Linear Strain Triangles
ξm = 0 = ξj
ξ i = 0.7 5 ξ i = 1 1 or i
10/14/08
ξ m = 0.75 ξ m = 0.5
74
Natural Coordinates for a Triangular Element (Area Coordinates)
Centroid: ξ i = ξ j= ξ m = 1/3 10/14/08
Linear Strain Triangles
75
Node 1 2 3 4 5 6
10/14/08
ξ1
ξ2
ξ3
1 0 0 0 1/2 1/2
0 1 0 1/2 0 1/2
0 0 1 1/2 1/2 0
Linear Strain Triangles
76
Shape Functions Ni
1
2
_________________________________________________
1 2 3 4 5 6
10/14/08
1 0 0 0 0 0
0 1 0 0 0 0
3
4
5
6
_________________________________________________
0 0 1 0 0 0
Linear Strain Triangles
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
77
Can we write shape functions directly? ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 1/2 at node 4 ξ 1 = 1/2 at node 5 ξ 1 = 0 at node 6
10/14/08
N1 = ξ 1 ( 2ξ 1-1 )
Linear Strain Triangles
78
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 0 at node 5 ξ 2 = 1/2 at node 6
10/14/08
N2 = ξ 2 ( 2ξ 2-1 )
Linear Strain Triangles
79
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 1/2 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 0 at node 6
10/14/08
N3 = ξ 3 ( 2ξ 3-1 )
Linear Strain Triangles
80
ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 0 at node 4 ξ 1 = 1/2 at node 5 ξ 1 = 1/2 at node 6
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 0 at node 5 ξ 2 = 1/2 at node 6
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 1/2 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 0 at node 6
N4 = 4 ξ 2 ξ 3 N5 = 4 ξ 1 ξ 3 N6 = 4 ξ 1 ξ 2
10/14/08
Linear Strain Triangles
81
ξ1 = s ξ2 = t ξ3 = 1 − s − t
10/14/08
Linear Strain Triangles
82
Step 1 -Select Element Type
x = a1 + a 2s + a 3 t + a4 s + a 5s t + a6 t 2
2
y = a 7 + a 8 s + a 9 t + a 10 s + a 11 s t + a 12 t 2
10/14/08
Linear Strain Triangles
2
83
Step 1 -Select Element Type
x N1 = y 0
10/14/08
0
N2
0
N3
0
N4
0
N5
0
N6
N1
0
N2
0
N3
0
N4
0
N5
0
Linear Strain Triangles
x1 y 1 x 2 0 y 2 N6 x6 y 6
84
Step 1 -Select Element Type ∂ N i ∂ N i ∂ ξ1 ∂ N i ∂ ξ 2 ∂ N i ∂ ξ 3 = + + ∂s ∂ ξ1 ∂ s ∂ ξ 2 ∂ s ∂ ξ 3 ∂ s ∂ Ni ∂ Ni ∂ Ni = − ∂s ∂ ξ1 ∂ ξ 3 ∂ Ni ∂ Ni ∂ Ni = − ∂t ∂ ξ2 ∂ ξ3 10/14/08
Linear Strain Triangles
85
Step 1 -Select Element Type N1 = ξ 1 ( 2ξ 1-1 ) N2 = ξ 2 ( 2ξ 2-1 ) N3 = ξ 3 ( 2ξ 3-1 ) N4 = 4 ξ 2 ξ 3 N5 = 4 ξ 1 ξ 3 N6 = 4 ξ 1 ξ 2
10/14/08
ξ1 = s ξ2 = t ξ3 = 1 − s − t
Linear Strain Triangles
86
Step 1 -Select Element Type N1 = s ( 2s - 1 ) N2 = t ( 2t - 1 ) N3 = (1 - s - t) ( 1 - 2s - 2t) N4 = 4 t (1 - s - t) N5 = 4 s (1 - s - t) N6 = 4 s t 10/14/08
Linear Strain Triangles
87
Step 2 -Select Displacement Function
u N1 = v 0
10/14/08
0
N2
0
N3
0
N4
0
N5
0
N6
N1
0
N2
0
N3
0
N4
0
N5
0
Linear Strain Triangles
u1 v 1 u 2 0 v 2 N6 u 6 v6
88
Step 3 -Define Stress/Strain and Strain/Displacement ∂u ε x ∂x ∂v { ε} = ε y = γ ∂y xy ∂u ∂v + ∂y ∂x { ε} = [ B]{ d} 10/14/08
Linear Strain Triangles
89
Chain Rule:
∂f ∂f ∂ x ∂f = + ∂s ∂ x ∂s ∂ y ∂f ∂f ∂ x ∂f = + ∂t ∂x ∂t ∂y 10/14/08
Linear Strain Triangles
∂y ∂s ∂y ∂t 90
Cramer’s Rule: ∂f ∂s ∂f ∂t
∂f = ∂x ∂x ∂s ∂x ∂t 10/14/08
∂y ∂s ∂y ∂t ∂y ∂s ∂y ∂t
∂x ∂s ∂x ∂t
∂f = ∂y ∂x ∂s ∂x ∂t
Linear Strain Triangles
∂f ∂s ∂f ∂t ∂y ∂s ∂y ∂t 91
Jacobian matrix:
∂x ∂s [ J] = ∂ x ∂ t 10/14/08
∂y ∂s ∂y ∂ t
Linear Strain Triangles
92
Strains in terms of an operator matrix:
∂( ) ε x ∂x { ε} = ε y = 0 γ xy ∂( ) ∂y 10/14/08
Linear Strain Triangles
0 ∂ ( ) u ∂y v ∂( ) ∂x 93
∂( ) 1 ∂y ∂ ( ) ∂y ∂ ( ) = − ∂x ∂s ∂t J ∂t ∂s ∂( ) 1 ∂x ∂ ( ) ∂ x ∂ ( ) = − ∂y ∂t ∂ s J ∂ s ∂t 10/14/08
Linear Strain Triangles
94
∂y ∂ ( ) ∂y ∂ ( ) − ∂t ∂s ∂s ∂t εx 1 0 εy = γ J ∂x ∂ ( ) ∂x ∂ ( ) xy − ∂s ∂t ∂t ∂ s
10/14/08
0 ∂x ∂ ( ) ∂x ∂ ( ) u − ∂s ∂t ∂t ∂s v ∂y ∂ ( ) ∂y ∂ ( ) − ∂t ∂s ∂s ∂t
Linear Strain Triangles
95
{ ε} = [ D ′] [ N ] { d} ∂y ∂t 1 [ D′] = J ∂x ∂s 10/14/08
∂ ( ) ∂y ∂ ( ) − ∂s ∂s ∂t 0 ∂ ( ) ∂x ∂ ( ) − ∂t ∂t ∂s
0 ∂x ∂s ∂y ∂t
Linear Strain Triangles
∂ ( ) ∂x − ∂t ∂t ∂ ( ) ∂y − ∂s ∂s
∂( ) ∂s ∂( ) ∂t 96
Step 4 -Derive Element Stiffness Matrix and Equations
[ k ] = ∫∫ [ B ] [ D ] [ B ] T
t dx dy
A
f ( x , y ) dx dy = ∫∫ A
f ( s , t ) J ds dt ∫∫ A
[ k ] = ∫∫ [ B ] [ D ] [ B ] T
t J ds dt
A
10/14/08
Linear Strain Triangles
97
{ fb } = ∫ ∫ [ N] { Xb } t J ds dt T
A
{ fs } = ∫L [ N] { T} t J dL T
10/14/08
Linear Strain Triangles
98
Need to use numerical integration to evaluate the element stiffness matrix.
∫∫f (ξ , ξ 1
A
10/14/08
n
2
(
, ξ 3 ) dA = ∑ Wi f ξ1i , ξ 2i , ξ 2i i =1
Linear Strain Triangles
)
99
1-point Gauss Formula
Wi = 1 ξ 1i = 1/3 ξ 2i = 1/3 ξ 3i = 1/3 10/14/08
Linear Strain Triangles
100
3-point Gauss Formula
Wi = 1/3 ξ 1i = 1/3, 1/6, 1/6 ξ 2i = 1/6, 1/3 , 1/6 ξ 3i = 1/6 , 1/6, 1/3 10/14/08
Linear Strain Triangles
101
3-point Gauss Formula
Wi = 1/3 ξ 1i = 1/2, 1/2, 0 ξ 2i = 1/2, 0, 1/2 ξ 3i = 0, 1/2, 1/2 10/14/08
Linear Strain Triangles
102
4-point Gauss Formula Wi = -0.56250 ξ 1i = 1/3 ξ 2i = 1/3 ξ 3i = 1/3 Wi = 0.52083 33333 33333 ξ 1i = 0.6, 0.2, 0.2 ξ 2i = 0.2, 0.6, 0.2 ξ 3i = 0.2, 0.2, 0.6 10/14/08
Linear Strain Triangles
103
6-point Gauss Formula Wi = 0.10995 17436 55322 ξ 1i = 0.81684 75729 80459 ξ 2i = 0.09157 62135 09771 ξ 3i = 0.09157 62135 09771 Wi = 0.22338 15896 78011 ξ 1i = 0.10810 30181 68070 ξ 2i = 0.44594 84909 15965 ξ 3i = 0.44594 84909 15965 10/14/08
Linear Strain Triangles
104
7-point Gauss Formula Wi = 0.12593 91805 44827 ξ 1i = 0.79742 69853 53087 ξ 2i = 0.10128 65073 23456 ξ 3i = 0.10128 65073 23456 Wi = 0.13239 41527 88506 ξ 1i = 0.47014 20641 05115 ξ 2i = 0.47014 20641 05115 ξ 3i = 0.05971 58717 89770 10/14/08
Linear Strain Triangles
Wi = 0.225 ξ 1i = 1/3 ξ 2i = 1/3 ξ 3i = 1/3 105
Pascal’s Triangle:
Degree of polynomial 0
1 x x x x x5 10/14/08
x4 y
y
x y 3
2
2
xy
2
x y
4
y xy
2
3
1
x y 2
x3 y2
2
xy
3
3
x2 y3
3
y
2
4
y
x y4
4
y5
Linear Strain Triangles
5
106
Degree of polynomial Number of terms n p
10/14/08
1
3
2
6
3
10
4
15
Linear Strain Triangles
107
n
φ = ∑ a i ξ 1q ξ r2 ξ s3 i =1
q+r+s =p Quadratic Triangle n=6 p=2 ( p + 1)( p + 2 ) n= 2 φ = a 1 ξ 12 + a 2 ξ 22 + a 3 ξ 23 + a 4 ξ 1 ξ 2 + a 5 ξ 2 ξ 3 + a 6 ξ 3 ξ 1 Equivalent to : φ = b 1 + b 2 x + b 3 y + b 4 x 2 + d 5 xy + b 6 y 2 10/14/08
Linear Strain Triangles
108
Nodal Numbering Schemes: 3
3 6
5
2 1
1
3 8
4
2
7
9 10
6 2
10/14/08
1
4Linear Strain 5Triangles
109
Quadratic Triangle φ = a 1 ξ 12 + a 2 ξ 22 + a 3 ξ 23 + a 4 ξ 1 ξ 2 + a 5 ξ 2 ξ 3 + a 6 ξ 3 ξ 1
Node 1 2 3 4 5 6 10/14/08
ξ1
ξ2
ξ3
1 0 0 1/2 0 1/2
0 1 0 1/2 1/2 0
0 0 1 0 1/2 1/2
Linear Strain Triangles
110
Shape Functions
Ni
1
2
_________________________________________________
1 2 3 4 5 6
10/14/08
1 0 0 0 0 0
0 1 0 0 0 0
3
4
5
6
_________________________________________________
0 0 1 0 0 0
Linear Strain Triangles
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
111
Can we write shape functions directly? ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 1/2 at node 4 ξ 1 = 0 at node 5 ξ 1 = 1/2 at node 6
10/14/08
N1 = ξ 1 ( 2ξ 1-1 )
Linear Strain Triangles
112
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 1/2 at node 5 ξ 2 = 0 at node 6
10/14/08
N2 = ξ 2 ( 2ξ 2-1 )
Linear Strain Triangles
113
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 1 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 1/2 at node 6
10/14/08
N3 = ξ 3 ( 2ξ 3-1 )
Linear Strain Triangles
114
ξ 1 = 1 at node 1 ξ 1 = 0 at node 2 ξ 1 = 0 at node 3 ξ 1 = 1/2 at node 4 ξ 1 = 0 at node 5 ξ 1 = 1/2 at node 6
ξ 2 = 0 at node 1 ξ 2 = 1 at node 2 ξ 2 = 0 at node 3 ξ 2 = 1/2 at node 4 ξ 2 = 1/2 at node 5 ξ 2 = 0 at node 6
ξ 3 = 0 at node 1 ξ 3 = 0 at node 2 ξ 3 = 1 at node 3 ξ 3 = 0 at node 4 ξ 3 = 1/2 at node 5 ξ 3 = 1/2 at node 6
N4 = 4 ξ 1 ξ 2 N5 = 4 ξ 2 ξ 3 N6 = 4 ξ 3 ξ 1
10/14/08
Linear Strain Triangles
115
∂u ∂N i εX = =∑ ui ∂x i =1 ∂x 3 ∂N i ∂N i ∂ξ j =∑ ∂x i = 1 ∂ ξ j ∂x 6
10/14/08
Linear Strain Triangles
116
Cubic Element
ξ i = 1, 0, 1/3, 2/3
N1 = 1/2 ξ 1(3 ξ 1 -1)(3 ξ 1 -2) N2 = 1/2 ξ 2(3 ξ 2 -1)(3 ξ 2 -2) N3 = 1/2 ξ 2(3 ξ 2 -1)(3 ξ 2 -2) N4 = 9/2 ξ 1 ξ 2(3 ξ 1-1) N5 = 9/2 ξ 2 ξ 1(3 ξ 2-1) N6 = 9/2 ξ 2 ξ 1(3 ξ 1-1) N7 = 9/2 ξ 2 ξ 3(3 ξ 2-1) N8 = 9/2 ξ 2 ξ 3(3 ξ 3-1) N9 = 9/2 ξ 1 ξ 3(3 ξ 1-1) N10 = 27 ξ 1 ξ 2 ξ 3 10/14/08
Linear Strain Triangles
117
