STAAD.Pro Knowledge Base

Issue #: SP-1794 Date Posted: 1/27/2002
Description: Do you have any information I can use in understanding instability messages?
Version: ALL Build No: ALL
Solution: Three questions and their answers are provided below to help explain this issue.

Question :

I have analyzed a structure and find that there are instability messages in the .anl (output) file, as follows :



***WARNING - INSTABILITY AT JOINT 26 DIRECTION = FX
PROBABLE CAUSE SINGULAR-ADDING WEAK SPRING
K-MATRIX DIAG= 5.3274384E+03 L-MATRIX DIAG= 0.0000000E+00 EQN NO 127
***NOTE - VERY WEAK SPRING ADDED FOR STABILITY

**NOTE** STAAD DETECTS INSTABILITIES AS EXCESSIVE LOSS OF SIGNIFICANT DIGITS
DURING DECOMPOSITION. WHEN A DECOMPOSED DIAGONAL IS LESS THAN THE
BUILT-IN REDUCTION FACTOR TIMES THE ORIGINAL STIFFNESS MATRIX DIAGONAL,
STAAD PRINTS A SINGULARITY NOTICE. THE BUILT-IN REDUCTION FACTOR
IS 1.000E-09

THE ABOVE CONDITIONS COULD ALSO BE CAUSED BY VERY STIFF OR VERY WEAK
ELEMENTS AS WELL AS TRUE SINGULARITIES.


What is the significance of such messages?

Answer :

An instability is a condition where a load applied on the structure is not able to make its way into the supports because no paths exist for the load to flow through, and may result in a lack of equilibrium between the applied load and the support reaction.

There is some explanation available in Section 1.18.1 of the STAAD.Pro Technical Reference Manual for the typical cause of instabilities. You will find it under the heading "Modeling and Numerical Instability Problems".

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Question :

If there are instability messages, does it mean my analysis results may be unsatisfactory?


Answer :

There are many situations where instabilities are unimportant and the STAAD approach of adding a weak spring is an ideal solution to the problem. For example, sometimes an engineer will release the MX torsion in a single beam or at the ends of a series of members such that technically the members are unstable in torsion. If there is no torque applied, this singularity can safely be "fixed" by STAAD with a weak torsional spring.

Similarly a column that is at a pinned support will sometimes be connected to members that all have releases such that they cannot transmit moments that cause torsion in the column. This column will be unstable in torsion but can be safely "fixed" by STAAD with a weak torsional spring.

Sometimes however, a section of a structure has members that are overly released to the point where that section can rotate with respect to the rest of the structure. In this case, if STAAD adds a weak spring, there may be large displacements because there are loads in the section that are in the direction of the extremely weak spring. Another way of saying it is, an applied load acts along an unstable degree of freedom, and causes excessive displacements at that degree of freedom.

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Question :

If there are instability messages, are there any simple checks to verify whether my analysis results are satisfactory?


Answer :

There are 2 important checks that should be carried out if instability messages are present.

a) A static equilibrium check. This check will tell us whether all the applied loading flowed through the model into the supports. A satisfactory result would require that the applied loading be in equilibrium with the support reactions.

b) The joint displacement check. This check will tell us whether the displacements in the model are within reasonable limits. If a load passes through a corresponding unstable degree of freedom, the structure will undergo excessive deflections at that degree of freedom.

One may use the PRINT STATICS CHECK option in conjunction with the PERFORM ANALYSIS command to obtain a report of both the results mentioned in the above checks. The STAAD output file will contain a report similar to the following, for every primary load case that has been solved for :


***TOTAL APPLIED LOAD ( KG METE ) SUMMARY (LOADING 1 )
SUMMATION FORCE-X = 0.00
SUMMATION FORCE-Y = -817.84
SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-
MX= 291.23 MY= 0.00 MZ= -3598.50


***TOTAL REACTION LOAD( KG METE ) SUMMARY (LOADING 1 )
SUMMATION FORCE-X = 0.00
SUMMATION FORCE-Y = 817.84
SUMMATION FORCE-Z = 0.00

SUMMATION OF MOMENTS AROUND THE ORIGIN-
MX= -291.23 MY= 0.00 MZ= 3598.50


MAXIMUM DISPLACEMENTS ( CM /RADIANS) (LOADING 1)
MAXIMUMS AT NODE
X = 1.00499E-04 25
Y = -3.18980E-01 12
Z = 1.18670E-02 23
RX= 1.52966E-04 5
RY= 1.22373E-04 23
RZ= 1.07535E-03 8


Go through these numbers to ensure that

i) The "TOTAL APPLIED LOAD" values and "TOTAL REACTION LOAD" values are equal and opposite.
ii) The "MAXIMUM DISPLACEMENTS" are within reasonable limits.