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


Outline

• Motivation and objective.

• Problem setup and analysis

• Implementation and results

• Conclusion

• Appendix


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.


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


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.


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 ��


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


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


�� 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.


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


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


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


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



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


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



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


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



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


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



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


−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


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


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.


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

Buckling Analysis Using RIKS Method

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