Download - Buckling Analysis Using RIKS Method
Term project by Dong Xue 1
Buckling Analysis Using Riks Method
Dong Xue
Term project for EM394g
Department of Engineering Mechanics
May,9th 2002
Term project by Dong Xue 2
Outline
• Motivation and objective.
• Problem setup and analysis
• Implementation and results
• Conclusion
• Appendix
Term project by Dong Xue 3
Motivation and objective
Static buckling analysis frequently involve structural
instabilities where the load-displacement response shows a
negative stiffness and the structure must release strain
energy to remain in equilibrium.
In such cases the load can reach a maximum sustainable
value and then can decrease. A typical response curve is
shown above. Zero stiffness (illustrated by the horizontal
tangent) poses a problem for the Newton-Raphson method
used in the regular STATIC analysis procedure because the
technique predicts an unbounded displacement increment,
often preventing further solution.
Term project by Dong Xue 4
One approach to analyze static buckling behavior is to use
the Riks method. This method is suitable when the
loading is proportional; that is, where the load
magnitudes are governed by a single scalar parameter.
The method can provide solutions even in cases of
complex, unstable response.
The goal of my project is
• Write a Fortran subroutine using Riks methods
• Build an one beam mode and a two beam model (both
with and without imperfection)
• Compute and plot configurations after deformed
• Compute and plot load as a function of displacement
Term project by Dong Xue 5
Problem setup and analysis
First, I built up a one beam model as following,
1 16
15
3217
2
27
3 2620
6
5 24
4 297
258
23
10 30
1114
28
313319
1812
9 13
22
21x
1.0
y
2.0 4.00 6.0 8.0 10.0
Wall
Pressure
The Beam model
The beam has 33 nodes and 5 elements. The left side of
the beam was fixed on a wall and there is a pressure on the
node 31. The whole beam was made of steel so the
Young’s module� � ��� � �����
and Poisson ration� � � ����
, see appendix bm01.model and bmx1.inp.
Then, I built up a two beam model with and without
imperfection, see the following figures.
Term project by Dong Xue 6
1200
1200
Two beam without imperfection
Two beam with imperfection
40
4
12
3
4
5
712
13
98
6
1517
16
1411
10
40
4
12
3
4
5
712
13
15
1617
98
6
1411
10
y
x
y
x
The model has 17 nodes and 2 elements. The node 4 and
node 15 are fixed. There is a pressures on node 7. The two
beam are also made of steel.
The Riks method finds static equilibrium at the end of each
increments, see the following graph, so that
����������� � � ����(1)
where is load,�
is equilibrium path and�
is
displacement. We consider only proportional loading ����
Term project by Dong Xue 7
so that
���� � � ������ �� (2)
with � being Load Proportionality Factor and� ��� �
being
the load pattern defined in the current RIKS step and� ����
satisfies the condition
���� �� � � �(3)
with�� � � �� � and
� � ���� � .
(u’ , F ’)
(u , F )
S
F
u
S
t equilibrium path
Since both loads and displacements are unknowns, the
concept of arc length � �is introduced. It is the distance
along the equilibrium solution path in load displacement
space. The tangent � � � ������ �� � � satisfies���� � � �� � . We
got���� � � �������� �
� � �. The initial guess
� � � � � � � to the
Term project by Dong Xue 8
next point is,
� � � � � � � �� �� � � � � � � �� � (4)
We can calculate successive iterates� � � � � � � � � � � � �������
which satisfying the chasfield constraint,
� ���� � � � � � � �� � � � � � � � � � �(5)
and
� � � � � � � � � � � (6)
With the above formulas we can get �
In cases of severe nonlinearity, the Riks method may not
converge or may converge to previously obtained
equilibrium configurations. Often such problems are caused
by the arc length being too large. we try to find the typical
arc lengths used and rerun the analysis with a smaller arc
length.
we require methods to identify when a step is complete. I
specify a arc length �� � � ��, a maximum value of path
iteration ���� ��� � ���, a maximum value of successive
iteration ���� � � ��� � ���, a minimum value of arc length
Term project by Dong Xue 9
�� � �� � � ���and an error tolerance � � � � � � � �
at
which the step will complete or the corresponding values
are less than the specifications. If neither of these
conditions is reached, routine will stop.
Term project by Dong Xue 10
Implementation
For one beam model, I output 33 nodes’ coordinates after
deformed and ouput load values with respect to the
corresponding displacements, using the fortran code. Then
I plot the deformed beam and the load as a function of
displacement, using matlab.
The beam without imperfection can be plots as,
0 2 4 6 8 10 12−0.5
0
0.5
1
1.5
x axis
y ax
is
Beam without imperfection
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0 0.5 1 1.5 2 2.5 30
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
displacement
load
Beam without imperfection
Load as a function of displacement
Term project by Dong Xue 12
Then I tried to get the 1st mode plots by adding the
imperfection dy = 0.01 to node 18 and node 20. The plots I
got are as followings,
0 2 4 6 8 10 120
0.5
1
1.5
2
2.5
3
3.5
4
x axis
y ax
is
Beam with inperfection dy = 0.01at node 18 and 20
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0 0.5 1 1.5 2 2.5
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
displacement
load
Beam with imperfection dy = 0.01 at node 18 and 20
Load as a function of displacement
Term project by Dong Xue 14
I also can get the 1st mode plots by adding the imperfection
dy = -0.01 to node 18 and node 20. The plots I got are as
followings,
0 2 4 6 8 10 12−3
−2.5
−2
−1.5
−1
−0.5
0
0.5
1
x axis
y ax
is
Beam with imperfection dy = −0.01 at node 18 and 20
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0 0.5 1 1.5 2 2.5 3
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
displacement
load
Beam with imperfection dy= 0.01 at node 18 and 20
Load as a function of displacement
Term project by Dong Xue 16
The 2nd mode plots can be got by adding the imperfection
dy = 0.001 to node 12 , node 14 and dy = -0.001 to node 30
, node 32. The plots I got are as followings,
0 2 4 6 8 10 12−0.5
0
0.5
1
1.5
2
x axis
y ax
is
Beam with imperfection dy = 0.001 at node 12 ,14 and dy = −0.01 at node 24 ,26
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0 0.5 1 1.5 2 2.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
displacement
load
Beam with imperfection dy = 0.001 at node 12, 14 and dy = − 0.001at node 24, 26
Load as a function of displacement
Term project by Dong Xue 18
In the same way, for two beam model I output 17 nodes’
coordinates after deformed and ouput load values with
respect to the corresponding displacements, using the
fortran code. Then I plot the deformed beam and the load
as a function of displacement, using matlab.
The two beams without imperfection can be plots as,
0 10 20 34.46 40 50 60 69.28
0
5
10
15
20
x axis
y ax
is
Two beam without imperfection
Term project by Dong Xue 19
0 5 10 15 20 25 30 35−8
−6
−4
−2
0
2
4
6
8
displace in y direction
load
Stress as a function of displacement in Y direction in point 7
Load as a function of displacement
Term project by Dong Xue 20
The two beams with imperfection can be plots as,
−2 0 10 20 30 34.6 40 51.96 60−10
−5
0
5
15
10
20
x axis
y ax
is
Two beams with imperfection
Term project by Dong Xue 21
−5 −4 −3 −2 −1 0 1−5
−4
−3
−2
−1
0
1
2
3
4
5
Displacement in X direciton
load
Stress as a function of displacement in X direction in point 7
Load as a function of displacement in X direction
Term project by Dong Xue 22
0 5 10 15 20 25 30−5
−4
−3
−2
−1
0
1
2
3
4
5
Displacement in Y direction
Load
Stress as a function of displacement in Y direction in point 7
Load as a function of displacement in Y direction
Term project by Dong Xue 23
Conclusion
In this project, we test one beam mode and two beam
mode(both with and without imperfection). We know the
Riks method is a good way to analyze static buckling
behavior.
Term project by Dong Xue 24
Appendix
• Models
• One beam model without imperfection bm01 model
• Two beams model without imperfection twbb model
• Two beams model with imperfectiontwba model
• Inputs
• For one beam bm01 inp
• For two beams twbb inp
• Codes
• Fortran90 code new solver f90
• Matlab code for one beam pic m
• Matlab code for two beams pic2 m