Fundamentals of Finite Element Analysis phần 5 docx
Fundamentals of Finite Element Analysis phần 5 docx
... 10-node
triangular element shown in Figure P6.13. Note that the nodes are equally spaced
on their edges.
1 2 3 4 5
x
x
1
ϭ 2
x
2
ϭ 2. 25
x
3
ϭ 2 .5
x
4
ϭ 2. 75
x
5
ϭ 3
Hutton: Fundamentals of
Finite Element Analysis
6. ... Solutions
Four Elements, Eight Elements,
x (inches) T (
◦
F) T (
◦
F)
0 180 180
0 .5 158 .08* 155 .31
1.0 136.16 136.48
1 .5 123 .59 * 122.19
2.0 111.0...
Fundamentals of Finite Element Analysis phần 1 pdf
... Basic Concepts of the
Finite Element Method
Text © The McGraw−Hill
Companies, 2004
1 .5 Examples of Finite Element Analysis 15
0.00197Љ
Z
X
0. 25
0.488Љ
Figure 1.9
Finite element model of a thin-walled
heat ... solutions.
36
Hutton: Fundamentals of
Finite Element Analysis
1. Basic Concepts of the
Finite Element Method
Text © The McGraw−Hill
Companies,...
Fundamentals of Finite Element Analysis phần 2 pps
... are
10
5
1.3 25 1.3 25 0 0 −1.3 25 −1.3 25
1.3 25 1.3 25 0 0 −1.3 25 −1.3 25
003. 75 0 −3. 75 0
000000
−1.3 25 −1.3 25 −3. 75 0 5. 0 75 1.3 25
−1.3 25 −1.3 25 0 0 1.3 25 1.3 25
ۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙ
ۘ
ۘ
ۘ
ۘ
ۘ
ۘ
ۘ
0
0
0
0
U
5
U
6
ۙ
=
F
1
F
2
F
3
F
4...
Fundamentals of Finite Element Analysis phần 3 pps
... resulting
5 × 5
system of equations, we obtain the results
1
= 9.3638(10
−4
) rad
v
2
=−0.73811 mm
2
=−0.009 253 8 rad
v
3
= 5. 552 3 mm
3
=−0.019444 rad
Hutton: Fundamentals of
Finite Element Analysis
4. ... shown,
determine
(b)
6 in. 6 in.
0.8 75 in.
t ϭ 0. 25 in.
10 lb
0.6 25 in.
(a)
12 in.
1 in. 0 .5 in.
t ϭ 0. 25 in.
10 lb
Hutton: Fundamentals of
Finite...
Fundamentals of Finite Element Analysis phần 4 pps
... node.
EXAMPLE 5. 5
Hutton: Fundamentals of
Finite Element Analysis
5. Method of Weighted
Residuals
Text © The McGraw−Hill
Companies, 2004
5. 4 Application of Galerkin’s Method to Structural Elements 151
Galerkin’s ... 04.40
T
1
T
2
T
3
T
4
=
11.31
11.31
11.31
363.31
Hutton: Fundamentals of
Finite Elemen...
Fundamentals of Finite Element Analysis phần 6 pptx
... −0. 258 5 0 0 0
0 −0.1003 0.6906 0 −0. 258 5 −0.0713 0 0 0
−0.0713 −0. 258 5 0 1.3812 −0.2006 0 −0.0713 −0. 258 5 0
−0. 258 5 −0.2006 −0. 258 5 −0.2006 2 .53 08 −0.2006 −0. 258 5 −0.2006 −0. 258 5
0 −0. 258 5 −0.0713 ... −0. 258 5
0 −0. 258 5 −0.0713 0 −0.2006 1.3812 0 −0. 258 5 −0.0713
000−0.0713 −0. 258 5 0 0.7484 −0. 258 5 0
000−0. 258 5 −0.2006 −0. 258 5 −0. 258 5 1.3812 −0.0713
0000−0. 2...
Fundamentals of Finite Element Analysis phần 7 pdf
... 99.37 47.109 45. 2 15
22 67.10 64.67 51 .53 5 49.121
23 40 .55 39.36 57 .836 55 .499
24 18.98 18.28 68.142 65. 4 25
45 67.88 65. 89 41.706 40.799
46 103.87 100.74 42. 359 41.018
This solution is actually ... solution; 40 elements. (b) Streamlines (
= constant) for finite element solution
of Example 8.1.
(b)
6
7
2
5
4
3
1
24
23
22
27
42
46
35
41
51
48
47
49
53
21
20
17
25...
Fundamentals of Finite Element Analysis phần 8 pptx
... N
3
∂z
=
25
2A
k
(2)
=
1
4A
0
25
− 25
[
0 25 25
]
+
1
4A
− 25
0
25
[
− 25 0 25
]
[k
(2)
] =
1
1 250
00 0
0 6 25 −6 25
0 −6 25 6 25
+
1
1 250
6 25 0 −6 25
000
−6 25 0 6 25
=
0 .50 −0 .5
00 .5 −0 .5
−0 .5 ... Solid Mechanics
or
k
(1)
=
1
1 250
6 25 −6 25 0
−6 25 6 25 0
000
+
1
1 250
00 0
0 6 25 −6 25...
Fundamentals of Finite Element Analysis phần 9 ppsx
... =
1
√
m
0.2 956 0. 657 5 0.6930
0.4289 0 .56 18 −0.7124
0.6064 −0. 355 0 0.0782
To verify Equation 10.109, we form the triple product
[
A
]
T
[M][A] =
1
m
0.2 956 0.4289 0.6064
0. 657 5 0 .56 18 −0. 355 0
0.6930 ... =
k
m
0.2 956 0.4289 0.6064
0. 657 5 0 .56 18 −0. 355 0
0.6990 −0.7124 0.0782
3 −20
−23−1
0 −11
×
0.2 956 0. 657 5 0.6990
0.4289 0 .56 18 −0.7124
0...
Fundamentals of Finite Element Analysis phần 10 doc
... 2 85
what is it, 279
Finite element, 2, 12
Finite element analysis (FEA).
See Finite element method (FEM)
Finite element formulation
axisymmetric heat transfer, 273–276
axisymmetric stress
analysis, ... loading?
0
0
6 in.
3 in.
0 .5
in.
150 lb
50 lb/in.
15 in.
40 in.
10,000 N
3 m
9 m
Hutton: Fundamentals of
Finite Element Analysis
Back Matter Appendix D:...
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