Fundamentals of Finite Element Analysis phần 3 pps
Fundamentals of Finite Element Analysis phần 3 pps
... 2,
EI
z
L
3
=
207(10
3
)(2 133 33 )
30 0
3
= 1 635 .6N/mm
Table 4.4 Element- Displacement Correspondence
Global Displacement Element 1 Element 2 Element 3
1 100
2 200
3 311
4 420
5 030
6 040
7 0 03
... 5
system of equations, we obtain the results
1
= 9 .36 38(10
−4
) rad
v
2
=−0. 738 11 mm
2
=−0.0092 538 rad
v
3
=−5.55 23 mm
3
=−0.019444 rad
Hutton: Fundamen...
Fundamentals of Finite Element Analysis phần 2 pps
... as
[
K
]
=
1 .32 5 1 .32 5 0 0 −1 .32 5 −1 .32 5
1 .32 5 1 .32 5 0 0 −1 .32 5 −1 .32 5
0 03. 75 0 3. 75 0
000000
−1 .32 5 −1 .32 5 3. 75 0 5.075 1 .32 5
−1 .32 5 −1 .32 5 0 0 1 .32 5 1 .32 5
10
5
lb/in.
Incorporating ... 3. 77(10
−4
) 37 71
3 −4.0(10
−4
) −4000
4 1 .33 (10
−4
) 133 3
5 5 .33 (10
−4
) 533 3
6 −5.67(10
−4
) −5657
7 2.67(10
−4
) 2667
8 4.0...
Fundamentals of Finite Element Analysis phần 4 pps
... =
1
2A
[(x
2
y
3
− x
3
y
2
) + (y
2
− y
3
)x +(x
3
− x
2
)y]
1
+ [(x
3
y
1
− x
1
y
3
) + (y
3
− y
1
)x +(x
1
− x
3
)y]
2
+ [(x
1
y
2
− x
2
y
1
) + (y
1
− y
2
)x +(x
2
− x
1
)y]
3
(6 .36 )
Given ... x
2
y
3
,
y
x
1 (0, 0)
2
(x
2
, 0)
3 (x
3
, y
3
)
Figure 6.9 Three-node
triangle having an element
coordinate system attached
to the element.
Hutton: F...
Fundamentals of Finite Element Analysis phần 9 ppsx
... rectangular element is
m
(e)
=
2.61 .30 .71 .3
1 .32 .61 .30 .7
0.71 .32 .61 .3
1 .30 .71 .32 .6
0000
0000
0000
0000
0000
0000
0000
0000
2.61 .30 .71 .3
1 .32 .61 .30 .7
0.71 .32 .61 .3
1 .30 .71 .32 .6
(10)
3
kg
We ... s)
2
dr ds
=
150(5)
16
r −
r
3
3
(1 − s)
3
3
(−1)
1
−1
=
150(5)
16
(7. 83) (10)
−6
32
9
=...
Fundamentals of Finite Element Analysis phần 1 pdf
... forces on element 1 are equal and opposite as required for
equilibrium.
Element 2
3k −3k
−3k 3k
U
2
U
3
=
3k −3k
−3k 3k
−
F
4k
+
3
4
␦
␦
=
−
3F
4k
−
3
4
k␦
3F
4k
+
3
4
k␦
=
f
(2)
2
f
(2)
3
which ... postprocessing.)
Hutton: Fundamentals of
Finite Element Analysis
1. Basic Concepts of the
Finite Element Method...
Fundamentals of Finite Element Analysis phần 5 docx
... s
j
)
=
√
3
3
3
−1
√
3
3
−1
2
+
−
√
3
3
3
−1
√
3
3
−1
2
+
√
3
3
3
−1
−
√
3
3
−1
2
+
−
√
3
3
3
−1
−
√
3
3
−1
2
= 5 .33 333 333
And, as again may be ... are
W
i
= 1.0, i = 1, 2
. Therefore,
1
−1
(r
2
− 3r + 7) dr = (1)[(0.57 735 03)
2
− 3( 0.57 735 03) + 7]
+ (1)[(−0.57 735 03)
2
− 3( −0.57 735 03) +...
Fundamentals of Finite Element Analysis phần 6 pptx
... }=
−2.9801 3. 2165 0 0 0
3. 0595 0 .31 4 3. 2165 0 0
0 3. 0595 0 .31 4 3. 2165 0
00 3. 0595 0 .31 4 3. 2165
000 3. 0595 3. 2950
T
1
T
2
T
3
T
4
T
5
=
3. 532 5 ... conductance matrix is
k
(e)
=
0. 632 7 −0.10 03 −0.2585 −0.10 03
−0.10 03 0. 632 7 −0.10 03 −0.2585
−0.2585 −0.10 03...
Fundamentals of Finite Element Analysis phần 7 pdf
... 75.184 80
2 0 0 1.9 63 0
8 0 0 38 . 735 38 .4
16 1 23. 63 122.17 40. 533 40.510
20 142.48 137 .40 44.9 03 42.914
21 100. 03 99 .37 47.109 45.215
22 67.10 64.67 51. 535 49.121
23 40.55 39 .36 57. 836 55.499
24 18.98 ... 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
2...
Fundamentals of Finite Element Analysis phần 8 pptx
... at Node 107 of Example 9.4
x
y
xy
Element 1 2049 .3 187 .36 118.4
Element 2 2149.4 31 5.59 91.89
Element 12 1987 .3 322.72 204. 13
Element 99 18 53. 8 186.88 37 8 .36
Average 2009.8 2 53. 14 198.19
and ... 9.4
x
y
xy
e
Element 1 25 53. 5 209.71 179.87 2475.8
Element 2 1922.7 35 1.69 43. 55 1774.8
Element 12 1827.5 264.42 154.44 1 731 .5
Element 99 2189.0...
Fundamentals of Finite Element Analysis phần 10 doc
... stress
analysis, 36 5 36 8
one-dimensional conduction with
convection, 227– 230
plane stress, 33 0 33 3
stream function, 299 30 0
torsion, 37 8
two-dimensional conduction with
convection, 236 –240
Finite element ... 3
matrix
|
A
|
=
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
(A.16)
defined as
|
A
|
= a
11
(a
22
a
33
−a
23
a
32
) − a
12
(a
21
a...
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