Steel design issue with AS4100

[Anonymous]

Hi guys,

Some colleagues have pointed out what seems to be an error with the 4100 steel module.

Problems:

(refer to beam #1 in model attached.)

1. When robot is choosing the critical section for the design check, it seems that it is not picking the internal forces at the worst location .  In beam 1 in the file, the critical section is reported at L=0.0.  This is obviously not the case.  If lateral buckling is switched off, the forces seem to be the correct ones located at the expected position. 
2. The lateral flange restraints are also not recognised if only top flange restraint is selected.  However, if the bottom flange AND the top flange are chosen as restrained, it seems to work.  Whilst this may be a work around,  how can the tension flange control the restrained length in any case?  And what about in the moment reversal case, when the bottom flange is actually unrestrained in compression?

Other issues not related to the attched model but reported:

1.  For some cases the phiMby is LESS than the phiMoy.  How can this ever occur in a memebr with no axial tension?
2. In lateral buckling check, the is an issue with real versus relative co-ordinates of restraints.  If lateral restraints are applied using the "relative" parameter, the effective length is reported as an impossibly short length, ie 0.4m.    The actual length to restraint is 2m. 

Tony

[Artur.Kosakowski · Accepted solution]

Problems:

(refer to beam #1 in model attached.)

1. When robot is choosing the critical section for the design check, it seems that it is not picking the internal forces at the worst location .  In beam 1 in the file, the critical section is reported at L=0.0.  This is obviously not the case.  If lateral buckling is switched off, the forces seem to be the correct ones located at the expected position. 
2. The lateral flange restraints are also not recognized if only top flange restraint is selected.  However, if the bottom flange AND the top flange are chosen as restrained, it seems to work.  Whilst this may be a work around,  how can the tension flange control the restrained length in any case?  And what about in the moment reversal case, when the bottom flange is actually unrestrained in compression?

 I don't feel like being the expert in using the Australian steel code but I'll do my best

- if you switch off LTB verification then the unrestrained length of flanges is no longer the factor and the check is done at the point where you have largest internal forces

- if you verify LTB then the capacity is reduced and for the set of forces you have along the beam the verification will return the same ratio in more then one point along the beam. In such case Robot shows the first point in which this ratio has been found (if you reverse the local coordinate system of the beam this would be the point which is now the end of the beam). You can test this forcing Robot to run verification at the end of the beam only (see the picture).

- if you restrain only the top flange then if the verification is done in the point where the bottom flange is under compression the entire length is assumed as unrestrained. To avoid this you may the settings shown on the attached picture.

Finally to avoid (most likely unwanted)  small negative bending at x=0L due to connection with other bars I would define the release there.

Other issues not related to the attched model but reported:

1.  For some cases the phiMby is LESS than the phiMoy.  How can this ever occur in a memebr with no axial tension?
2. In lateral buckling check, the is an issue with real versus relative co-ordinates of restraints.  If lateral restraints are applied using the "relative" parameter, the effective length is reported as an impossibly short length, ie 0.4m.    The actual length to restraint is 2m.

I need the sample model to try to answer these questions.

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[Anonymous]

Artur,  thanks for quick reply..

Model was attached last post

Tony

[Artur.Kosakowski]

I used the model you attached to answer on problems but I need models to try to answer on "Other issues not related to the attched model".

[Anonymous]

oops. yes of course

[Anonymous]

Hi guys,

seems to me the AS4100 module has the compression and tension flanges confused..............

[Artur.Kosakowski]

IMHO it is correct. The inactive is the flange under tension which is governed by the sign of the bending moment. You can make a quick test changing the value of bending in the manual check dialog (see the attached picture).

If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.

[Anonymous]

I've just got hold of the other model with the other problems referred to in the original post. 

In this example, the beam #9 has a phiMby smaller than the phiMoy, we're unsure when this could occur as phiMoy is Mby reduced by axial compression

Model attached.  Beam 9 to investigate

Tony

[Anonymous]

bump

[Artur.Kosakowski]

Tony,

Give me a chance please. I need to learn AS code first (in case you don't know I do not use it on daily basis) 

[Artur.Kosakowski · Accepted solution]

Phi*Moy is in this case calculated according to the point 8.4.4.1 Compression members. This is due to the small compression force shown on the attached picture.

For H (or I ) sections that fulfill the required criteria Robot uses the alternative method from the page 92 of AS code (see the picture). According to this method the upper limit for Moy is Mry rather than Mby.

You can exclude small values of axial forces in the way shown on the attached picture too.

BTW: in this case Phi*Moy is sort of 'excessively' shown as it is not used in the bar's verification.

If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.

[Anonymous]

Artur,

Thanks a lot for all your help answering my endless questions.......................

(and you said you don't know alot of AS4100) 

Tony

[Artur.Kosakowski]

No problem. I learn myself as well .