examples from FEM
30
CHAPTER SIX Plates and Shells à Example 6.3 Square plate subjected to distributed load (p. 302) Consider solution of a square plate subjected to a uniformly distributed load. Two opposite sides of the plate are clamped and the other two are simply supported. Use the following numerical data. 1 µ 1 plate h = 0.1 E = 10920 n = 0.3 q = 1 Taking advantage of symmetry we model a 1/4 of the plate using only four elements as shown in Figure 6.12. The element side 1-4 is clamped, side 1-2 is simply supported, and the other two sides have symmetry boundary condition. Thus the following essential boundary conditions are imposed. Side1 - 4 : w = 0 q x = 0 q y = 0 Side1 - 2 : w = 0 q y = 0 Side 2 - 3 : q y = 0 Side 3 - 4 : q x = 0 1
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2D implementation of finite elements
Transcript of examples from FEM
-
CHAPTER SIX
Plates and Shells
Example 6.3 Square plate subjected to distributed load (p. 302)Consider solution of a square plate subjected to a uniformly distributed load. Two opposite sides of the plate are clampedand the other two are simply supported. Use the following numerical data.
1 1 plate h = 0.1 E = 10920 n = 0.3 q = 1
Taking advantage of symmetry we model a 1/4 of the plate using only four elements as shown in Figure 6.12. The elementside 1-4 is clamped, side 1-2 is simply supported, and the other two sides have symmetry boundary condition. Thus thefollowing essential boundary conditions are imposed.
Side1 - 4 : w = 0 qx = 0 qy = 0
Side1 - 2 : w = 0 qy = 0
Side 2 - 3 : qy = 0
Side 3 - 4 : qx = 0
1
-
0 0.50.50 0.25x
00
0.50.5
0.25
y
1 2
34
51
2
3
4
Figure 6.12. Four triangular plate element model
Global equations at start of the element assembly process
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
w1HqxL1HqyL1w2HqxL2HqyL2w3HqxL3HqyL3w4HqxL4HqyL4w5HqxL5HqyL5
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
=
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
000000000000000
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzEquations for element 1
Element nodes: 1 ikjjj 00 y{zzz 2 ikjjjjj1414
y{zzzzz 3 ikjjjj 012 y{zzzzh = 0.1 E = 10920 n = 0.3 D = 1. q = 1.
Mapping to the master element
xHs,tL = s4
yHs,tL = s4
+ t2
detJ = 18
2
-
Gauss quadrature points and weights
Point Weight
1s 13t 13
- 932
2s 15t 15
2596
3s 35t 15
2596
4s 15t 35
2596
After summing contributions from all points, the element equations as follows:
k =
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. 7.26667 -4.2 -48. 3.4 -4.2 0 1.33333 -3.67.26667 1.7 -0.65 -7.93333 -0.108333 -0.341667 0.666667 0.0583333 -0.991667-4.2 -0.65 0.733333 5.8 -0.0583333 0.275 -1.6 0.0583333 0.441667-48. -7.93333 5.8 96. 4. 8.4 -48. 3.93333 9.83.4 -0.108333 -0.0583333 4. 2.1 -0.05 -7.4 0.708333 1.10833-4.2 -0.341667 0.275 8.4 -0.05 1.6 -4.2 0.391667 0.2250 0.666667 -1.6 -48. -7.4 -4.2 48. -5.26667 -6.21.33333 0.0583333 0.0583333 3.93333 0.708333 0.391667 -5.26667 0.883333 0.533333-3.6 -0.991667 0.441667 9.8 1.10833 0.225 -6.2 0.533333 1.78333
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzrT = H 0.0208333 0.00208333 -0.000520833 0.0208333 -0.000260417 0.00130208 0.0208333 -0.00182292i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. 7.26667 -4.2 -48. 3.4 -4.2 2.22045 10-15 1.33333 -3.67.26667 1.7 -0.65 -7.93333 -0.108333 -0.341667 0.666667 0.0583333 -0.9-4.2 -0.65 0.733333 5.8 -0.0583333 0.275 -1.6 0.0583333 0.44-48. -7.93333 5.8 96. 4. 8.4 -48. 3.93333 9.83.4 -0.108333 -0.0583333 4. 2.1 -0.05 -7.4 0.708333 1.10-4.2 -0.341667 0.275 8.4 -0.05 1.6 -4.2 0.391667 0.221.33227 10-15 0.666667 -1.6 -48. -7.4 -4.2 48. -5.26667 -6.21.33333 0.0583333 0.0583333 3.93333 0.708333 0.391667 -5.26667 0.883333 0.53-3.6 -0.991667 0.441667 9.8 1.10833 0.225 -6.2 0.533333 1.78
The element contributes to 81, 2, 3, 13, 14, 15, 10, 11, 12< global degrees of freedom.Adding element equations into appropriate locations we have
3
-
ik
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. 7.26667 -4.2 0 0 0 0 0 0 0 1.33333 -3.6 -48. 3.47.26667 1.7 -0.65 0 0 0 0 0 0 0.666667 0.0583333 -0.991667 -7.93333 -0.108333-4.2 -0.65 0.733333 0 0 0 0 0 0 -1.6 0.0583333 0.441667 5.8 -0.05833330 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 00 0.666667 -1.6 0 0 0 0 0 0 48. -5.26667 -6.2 -48. -7.41.33333 0.0583333 0.0583333 0 0 0 0 0 0 -5.26667 0.883333 0.533333 3.93333 0.708333-3.6 -0.991667 0.441667 0 0 0 0 0 0 -6.2 0.533333 1.78333 9.8 1.10833-48. -7.93333 5.8 0 0 0 0 0 0 -48. 3.93333 9.8 96. 4.3.4 -0.108333 -0.0583333 0 0 0 0 0 0 -7.4 0.708333 1.10833 4. 2.1-4.2 -0.341667 0.275 0 0 0 0 0 0 -4.2 0.391667 0.225 8.4 -0.05
Equations for element 2
Element nodes: 1 ikjjj 00 y{zzz 2 ikjjjj 120 y{zzzz 3 ikjjjjj1414
y{zzzzzh = 0.1 E = 10920 n = 0.3 D = 1. q = 1.
Mapping to the master element
xHs,tL = s2
+ t4
yHs,tL = t4detJ = 18
Gauss quadrature points and weights
Point Weight
1s 13t 13
- 932
2s 15t 15
2596
3s 35t 15
2596
4s 15t 35
2596
After summing contributions from all points, the element equations as follows:
4
-
k =
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. 6.2 -5.26667 0 1.6 0.666667 -48. 4.2 -7.46.2 1.78333 -0.533333 3.6 0.441667 0.991667 -9.8 0.225 -1.10833-5.26667 -0.533333 0.883333 1.33333 -0.0583333 0.0583333 3.93333 -0.391667 0.7083330 3.6 1.33333 48. 4.2 7.26667 -48. 4.2 3.41.6 0.441667 -0.0583333 4.2 0.733333 0.65 -5.8 0.275 0.05833330.666667 0.991667 0.0583333 7.26667 0.65 1.7 -7.93333 0.341667 -0.108333-48. -9.8 3.93333 -48. -5.8 -7.93333 96. -8.4 4.4.2 0.225 -0.391667 4.2 0.275 0.341667 -8.4 1.6 0.05-7.4 -1.10833 0.708333 3.4 0.0583333 -0.108333 4. 0.05 2.1
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzrT = H 0.0208333 0.00078125 -0.00182292 0.0208333 0.000520833 0.00208333 0.0208333 -0.00130208 -0i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. 6.2 -5.26667 -3.55271 10-15 1.6 0.666667 -48. 4.26.2 1.78333 -0.533333 3.6 0.441667 0.991667 -9.8 0.225-5.26667 -0.533333 0.883333 1.33333 -0.0583333 0.0583333 3.93333 -0.391667-4.44089 10-15 3.6 1.33333 48. 4.2 7.26667 -48. 4.21.6 0.441667 -0.0583333 4.2 0.733333 0.65 -5.8 0.2750.666667 0.991667 0.0583333 7.26667 0.65 1.7 -7.93333 0.341667-48. -9.8 3.93333 -48. -5.8 -7.93333 96. -8.44.2 0.225 -0.391667 4.2 0.275 0.341667 -8.4 1.6-7.4 -1.10833 0.708333 3.4 0.0583333 -0.108333 4. 0.05
The element contributes to 81, 2, 3, 4, 5, 6, 13, 14, 15< global degrees of freedom.Adding element equations into appropriate locations we havei
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
96. 13.4667 -9.46667 0 1.6 0.666667 0 0 0 0 1.33333 -3.613.4667 3.48333 -1.18333 3.6 0.441667 0.991667 0 0 0 0.666667 0.0583333 -0.991-9.46667 -1.18333 1.61667 1.33333 -0.0583333 0.0583333 0 0 0 -1.6 0.0583333 0.441660 3.6 1.33333 48. 4.2 7.26667 0 0 0 0 0 01.6 0.441667 -0.0583333 4.2 0.733333 0.65 0 0 0 0 0 00.666667 0.991667 0.0583333 7.26667 0.65 1.7 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 00 0.666667 -1.6 0 0 0 0 0 0 48. -5.26667 -6.21.33333 0.0583333 0.0583333 0 0 0 0 0 0 -5.26667 0.883333 0.53333-3.6 -0.991667 0.441667 0 0 0 0 0 0 -6.2 0.533333 1.78333-96. -17.7333 9.73333 -48. -5.8 -7.93333 0 0 0 -48. 3.93333 9.87.6 0.116667 -0.45 4.2 0.275 0.341667 0 0 0 -7.4 0.708333 1.10833-11.6 -1.45 0.983333 3.4 0.0583333 -0.108333 0 0 0 -4.2 0.391667 0.225
Equations for element 3
Element nodes: 1 ikjjjj 120 y{zzzz 2 ikjjjjj1212
y{zzzzz 3 ikjjjjj1414
y{zzzzzh = 0.1 E = 10920 n = 0.3 D = 1. q = 1.
Mapping to the master element
5
-
xHs,tL = s2
+ 12
H-s - t + 1L + t4
yHs,tL = s2
+ t4
detJ = 18Gauss quadrature points and weights
Point Weight
1s 13t 13
- 932
2s 15t 15
2596
3s 35t 15
2596
4s 15t 35
2596
After summing contributions from all points, the element equations as follows:
k =
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. 5.26667 6.2 0 -0.666667 1.6 -48. 7.4 4.25.26667 0.883333 0.533333 -1.33333 0.0583333 0.0583333 -3.93333 0.708333 0.3916676.2 0.533333 1.78333 3.6 -0.991667 0.441667 -9.8 1.10833 0.2250 -1.33333 3.6 48. -7.26667 4.2 -48. -3.4 4.2-0.666667 0.0583333 -0.991667 -7.26667 1.7 -0.65 7.93333 -0.108333 -0.3416671.6 0.0583333 0.441667 4.2 -0.65 0.733333 -5.8 -0.0583333 0.275-48. -3.93333 -9.8 -48. 7.93333 -5.8 96. -4. -8.47.4 0.708333 1.10833 -3.4 -0.108333 -0.0583333 -4. 2.1 -0.054.2 0.391667 0.225 4.2 -0.341667 0.275 -8.4 -0.05 1.6
rT = H 0.0208333 0.00182292 0.00078125 0.0208333 -0.00208333 0.000520833 0.0208333 0.000260417 -0i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. 5.26667 6.2 1.59872 10-14 -0.666667 1.6 -48. 7.4 4.25.26667 0.883333 0.533333 -1.33333 0.0583333 0.0583333 -3.93333 0.708333 0.3916.2 0.533333 1.78333 3.6 -0.991667 0.441667 -9.8 1.10833 0.2251.5099 10-14 -1.33333 3.6 48. -7.26667 4.2 -48. -3.4 4.2-0.666667 0.0583333 -0.991667 -7.26667 1.7 -0.65 7.93333 -0.108333 -0.341.6 0.0583333 0.441667 4.2 -0.65 0.733333 -5.8 -0.0583333 0.275-48. -3.93333 -9.8 -48. 7.93333 -5.8 96. -4. -8.47.4 0.708333 1.10833 -3.4 -0.108333 -0.0583333 -4. 2.1 -0.054.2 0.391667 0.225 4.2 -0.341667 0.275 -8.4 -0.05 1.6
The element contributes to 84, 5, 6, 7, 8, 9, 13, 14, 15< global degrees of freedom.Adding element equations into appropriate locations we have
6
-
ik
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
96. 13.4667 -9.46667 0 1.6 0.666667 0 0 013.4667 3.48333 -1.18333 3.6 0.441667 0.991667 0 0 0-9.46667 -1.18333 1.61667 1.33333 -0.0583333 0.0583333 0 0 00 3.6 1.33333 96. 9.46667 13.4667 0 -0.666667 1.61.6 0.441667 -0.0583333 9.46667 1.61667 1.18333 -1.33333 0.0583333 0.05833330.666667 0.991667 0.0583333 13.4667 1.18333 3.48333 3.6 -0.991667 0.4416670 0 0 0 -1.33333 3.6 48. -7.26667 4.20 0 0 -0.666667 0.0583333 -0.991667 -7.26667 1.7 -0.650 0 0 1.6 0.0583333 0.441667 4.2 -0.65 0.7333330 0.666667 -1.6 0 0 0 0 0 01.33333 0.0583333 0.0583333 0 0 0 0 0 0-3.6 -0.991667 0.441667 0 0 0 0 0 0-96. -17.7333 9.73333 -96. -9.73333 -17.7333 -48. 7.93333 -5.87.6 0.116667 -0.45 11.6 0.983333 1.45 -3.4 -0.108333 -0.0583333-11.6 -1.45 0.983333 7.6 0.45 0.116667 4.2 -0.341667 0.275
Equations for element 4
Element nodes: 1 ikjjjjj1212
y{zzzzz 2 ikjjjj 012 y{zzzz 3 ikjjjjj1414
y{zzzzzh = 0.1 E = 10920 n = 0.3 D = 1. q = 1.
Mapping to the master element
xHs,tL = 12
H-s - t + 1L + t4
yHs,tL = s2
+ 12
H-s - t + 1L + t4
detJ = 18Gauss quadrature points and weights
Point Weight
1s 13t 13
- 932
2s 15t 15
2596
3s 35t 15
2596
4s 15t 35
2596
After summing contributions from all points, the element equations as follows:
7
-
k =
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. -6.2 5.26667 0 -1.6 -0.666667 -48. -4.2 7.4-6.2 1.78333 -0.533333 -3.6 0.441667 0.991667 9.8 0.225 -1.108335.26667 -0.533333 0.883333 -1.33333 -0.0583333 0.0583333 -3.93333 -0.391667 0.7083330 -3.6 -1.33333 48. -4.2 -7.26667 -48. -4.2 -3.4-1.6 0.441667 -0.0583333 -4.2 0.733333 0.65 5.8 0.275 0.0583333-0.666667 0.991667 0.0583333 -7.26667 0.65 1.7 7.93333 0.341667 -0.108333-48. 9.8 -3.93333 -48. 5.8 7.93333 96. 8.4 -4.-4.2 0.225 -0.391667 -4.2 0.275 0.341667 8.4 1.6 0.057.4 -1.10833 0.708333 -3.4 0.0583333 -0.108333 -4. 0.05 2.1
rT = H 0.0208333 -0.00078125 0.00182292 0.0208333 -0.000520833 -0.00208333 0.0208333 0.00130208i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
48. -6.2 5.26667 2.04281 10-14 -1.6 -0.666667 -48. -4.2 7.4-6.2 1.78333 -0.533333 -3.6 0.441667 0.991667 9.8 0.225 -1.5.26667 -0.533333 0.883333 -1.33333 -0.0583333 0.0583333 -3.93333 -0.391667 0.702.04281 10-14 -3.6 -1.33333 48. -4.2 -7.26667 -48. -4.2 -3.-1.6 0.441667 -0.0583333 -4.2 0.733333 0.65 5.8 0.275 0.05-0.666667 0.991667 0.0583333 -7.26667 0.65 1.7 7.93333 0.341667 -0.-48. 9.8 -3.93333 -48. 5.8 7.93333 96. 8.4 -4.-4.2 0.225 -0.391667 -4.2 0.275 0.341667 8.4 1.6 0.057.4 -1.10833 0.708333 -3.4 0.0583333 -0.108333 -4. 0.05 2.1
The element contributes to 87, 8, 9, 10, 11, 12, 13, 14, 15< global degrees of freedom.Adding element equations into appropriate locations we havei
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
96. 13.4667 -9.46667 0 1.6 0.666667 0 0 013.4667 3.48333 -1.18333 3.6 0.441667 0.991667 0 0 0-9.46667 -1.18333 1.61667 1.33333 -0.0583333 0.0583333 0 0 00 3.6 1.33333 96. 9.46667 13.4667 0 -0.666667 1.61.6 0.441667 -0.0583333 9.46667 1.61667 1.18333 -1.33333 0.0583333 0.05833330.666667 0.991667 0.0583333 13.4667 1.18333 3.48333 3.6 -0.991667 0.4416670 0 0 0 -1.33333 3.6 96. -13.4667 9.466670 0 0 -0.666667 0.0583333 -0.991667 -13.4667 3.48333 -1.183330 0 0 1.6 0.0583333 0.441667 9.46667 -1.18333 1.616670 0.666667 -1.6 0 0 0 0 -3.6 -1.333331.33333 0.0583333 0.0583333 0 0 0 -1.6 0.441667 -0.0583333-3.6 -0.991667 0.441667 0 0 0 -0.666667 0.991667 0.0583333-96. -17.7333 9.73333 -96. -9.73333 -17.7333 -96. 17.7333 -9.733337.6 0.116667 -0.45 11.6 0.983333 1.45 -7.6 0.116667 -0.45-11.6 -1.45 0.983333 7.6 0.45 0.116667 11.6 -1.45 0.983333
Essential boundary conditions
8
-
Node dof Value
1w1HqxL1HqyL1
000
2 w2 02 HqyL2 03 HqxL3 03 HqyL3 04
w4HqxL4HqyL4000
Remove 81, 2, 3, 4, 6, 8, 9, 10, 11, 12< rows and columns.After adjusting for essential boundary conditions we haveikjjjjjjjjjjjjjjjjjjj
1.61667 -1.33333 -9.73333 0.983333 0.45-1.33333 96. -96. -7.6 11.6-9.73333 -96. 384. 0 00.983333 -7.6 0 7.4 00.45 11.6 0 0 7.4
y{zzzzzzzzzzzzzzzzzzzikjjjjjjjjjjjjjjjjjjjHqxL2w3
w5HqxL5HqyL5y{zzzzzzzzzzzzzzzzzzz =
ikjjjjjjjjjjjjjjjjjjj
0.002343750.04166670.083333300
y{zzzzzzzzzzzzzzzzzzz
Solving the final system of global equations we get8HqxL2 = 0.00962676, w3 = 0.00208276, w5 = 0.000981715, HqxL5 = 0.000859818, HqyL5 = -0.00385027
