steel design query

RE: steel design query

farhat sheen — Sun, 13 Jun 2021 06:01:37 GMT

Dear Suresh,

I couldn't find any slenderness check (kl/r <200 for compression) for fluxural members in any code or textbook.

If I use Main 1 for Sub beams, Main Beams and web tappered Rafters to bypass slenderness check, is it okay ?

Can you tell for which member and when we should use Main 1 parameter?

RE: steel design query

Sye — Thu, 10 Jun 2021 19:10:39 GMT

Seems like this thread is quite old. Answering the last question. Although flexural members may not be subjected to significant axial forces but design codes usually require flexural members to satisfy the axial check as well as interaction of axial + bending checks. Any steel member designed in STAAD.Pro is therefore checked for both axial and axial+flexure interaction. You would need to specify Ly Lz etc. unless the default lengths for the analytical members represent the correct unsupported lengths that should be considered for the flexural buckling calculations. MAIN 1 should only be used if you want to bypass the slenderness check for some reason. The software would still check the axial capacity as required by the code and hence you should always specify Ly Lz irrespective of whether you want the slenderness check to be by-passed or not.

RE: steel design query

farhat sheen — Thu, 10 Jun 2021 07:35:28 GMT

Dear Suresh,

As far as I know, the Lx, Ly, & Lz value will be used to calculate the Kl/r ratio. In the current case, we do not check slenderness for flexural members. So from my point of view, there shouldn't be any assignment of Lx, Lz, Ly for the beams.

If anyone assigns Lx, Ly, Lz for beams then they should assign 'Main1' as well. In the end, the result will be similar if you don't assign L values and if you assign L values with Main 1 for beams.

Can you solve if I'm incorrect. ?

Re: steel design query

Jungmo Ko — Wed, 20 Jun 2012 16:50:49 GMT

I am REALLY AND FULLY appreciated you. Cuz now, I could understand that Kx, Ky, Lx, Ly, UNT, UNB command.

Thank you so much!

Re: steel design query

Arunkumar Srinivasan — Thu, 26 May 2011 07:43:54 GMT

Dear Mr. Arun,

I believe either one should be sufficient. If you use both, the last one specified will be used in design...

Arun

Re: steel design query

sureshprsharma — Wed, 25 May 2011 20:24:57 GMT

I have no doubt about the capability of Arun. He has been of great help to me in the past. I am some how not convinced in the present case. Arun has mentioned in the first paragraph of his posting that top flange of the ISMB i.e. main beam will match with the top flange of ISMC. I understand that by matching Arun means that top flange of ISMB and ISMC will be at one level. Naturally the bottom flange will be at different level.This can only provide restraint to the main beam against Mx and My. This will not affect Mz of the main beam. In the UNL clear distance which is along the local x axis of the member should be input because that is the length which will play the key role in determining the Mz and design of beam. Even the extract from American code that you have posted states UNT is the face of the beam i.e.top facing the positive Y axis and UNT is the face of beam i.e. bottom facing the negative Y axis. Unless there is some support or restraint on the top and or bottom of the beam UNL will continue to be the face to face distance between the supports.

This is my view. Contradictions if any will be welcome and an addition to my knowledge.

Re: steel design query

Arun Subramanian — Wed, 25 May 2011 19:19:21 GMT

Dear Suresh,

I'm inclined to agree with Arun in this regard.

The parameters UNB/UNT (in case of Steel design design as per AISC) or UNL (for IS Code) is to specify the unrestrained length of compression flange to calculate effects of Lateral Torsional Buckling.

Refer Note 5 in Section 2.4 of the Staad Reference Manual/Help. (See attached PDF)

Dear Arun,

My question would be this. For the specified case (If the main beam has hogging moment), do we need to specify both UNB,UNT/UNL parameters and LY,LZ Parameters?

Or either one of them would be enough?

Regards,

Arun.

Re: steel design query

Arunkumar Srinivasan — Tue, 24 May 2011 18:48:50 GMT

Dear Sir,

The compression flange is not being restrained from developping stresses - it is being restrained against buckling. If primary member is ISMB450 and secondary member is ISMC75, only top flanges will match. Bottom flanges will not match. Hence, top flange will be restrained against buckling at that point but bottom flange will not be restrained...

Regarding effective length factor, similar tables are there in RCC column design also, see Annex E of IS456... There is no need for effective length factor in beams in RCC design.

Arun

Re: steel design query

sureshprsharma — Tue, 24 May 2011 18:09:30 GMT

Dear Arun ,

I am not able to understand as to how the compression flange alone can be restrained from developing compression stress in the top fibre while the tension flange will remain unrestrained and independently can take tension stress in the bottom fibre. Both compression and tension flange will have to work together otherwise there will be instability in the axial direction. One more thing specially from Mr. Arun. In your earlier posting you mentioned about the factor 0.85. whether this factor is applicable to steel structure alone or it can be for RCC structure as well particularly in case of beam or both for beam and column.

Re: steel design query

Arunkumar Srinivasan — Tue, 24 May 2011 14:13:14 GMT

Dear Sir,

As I understand, the unsupported length is to be provided for beams in calculating the effective length of the compression flange. If the compression flange is restrained by beams and/or slabs, we get higher allowable bending stresses. Hence, as I understand, we can take reduced length if there are cross beams that restrain the compression flange.

Please correct me if I am wrong.

Arun

Re: steel design query

sureshprsharma — Tue, 24 May 2011 13:53:38 GMT

Dear Arun,

According to me the UNL is the unsupported length of primary beam which is not supported/restrained in Y direction. The primary beam is restrained in Z direction by the secondarybeams in this case.Hence I feel that UNL of 10 meter has to be assigne to all the small beams forming part of the main beam irrespective of whether bottom flange or top flanges are connected because entire 10 meter length is going to behave as one beam having concentrated loads from secondary beams unrestrained in Y direction.. otherwise every piece of beam will be designed as independent beam for node to node distance of 2 m.

Re: steel design query

Arunkumar Srinivasan — Tue, 24 May 2011 12:37:47 GMT

Dear Sirs,

If there are secondary beams connected to match top of the primary beam and the primary beam is subjected to sagging moment, compression flange will be the top flange which will be restrained by the secondary beams. In that case, there is no need to send UNL or such parameter at all since effective length can be taken as the segment length.

If the secondary beam is at bottom level or the primary beam has hogging moment, then you need to use the parameters.

Arun

Re: steel design query

sureshprsharma — Tue, 24 May 2011 07:41:52 GMT

In IS 800 the command is UNL in lieu of both UNT and UNB. In the present case that you have cited you will have to assign UNL as 10 m for all the 5 pieces of beam which are of 2 meter between node to node and forming a part of 10 meter long beam.

Re: steel design query

Arun Subramanian — Mon, 23 May 2011 22:15:36 GMT

Dear Suresh,

Thanks for your reply. But it'll be much helpful to me if you can add some more clarity to the message.

I have a 10 m long beam(W16x40). It's split into 5 parts due to the secondary beams that are framing into the same. Now in the design parameters section, should I be using LZ = 10?

Or should I use UNB = 10? What difference will the two make?

As per the explanation of Design Parameters in STAAD, UNB, UNT parameters are used for calculating bending stresses for behavior like a beam. LY, LZ parameters are used for specifying effective lengths i.e., behavior similar to a column. Can these two be used interchangeably?

Kindly correct me if I'm wrong about the usage of these parameters.

Regards,

Arun.

Re: steel design query

Arunkumar Srinivasan — Sat, 21 May 2011 12:11:08 GMT

Dear Sir,

Both new and old code refers to steel sections. From first post it was mentioned as steel section, so I have been talking with steel only in mind... In case of concrete also ELY and ELZ are applicable instead of KY and KZ but only for columns.

Arun

Re: steel design query

sureshprsharma — Sat, 21 May 2011 06:44:08 GMT

Dear Arun,

A lot many thanks to you for the quick response. You have referred to Table 11 of the new code. I suppose by new code you mean IS 456 - 2000. But Table 11 of the above code refers to Characteristic compressive strength of concrete. It does consist of the factor i.e. 0.85 that you have stated. Please also clarify whether the factor 0.85 is applicable to the frame members or it is meant only individual beam and individual column. Please.

Re: steel design query

Arunkumar Srinivasan — Sat, 21 May 2011 06:02:17 GMT

Dear Sir,

The factor 0.85 would be applicable to the 10m beam. Hence, the effective length would be 0.85 x 10 or 8.5 m. Since we are dividing the beam into 4m component and 6 m component, if we do not give Ky/Kz/LY/LZ factors, their effective lengths would have been 4m and 6m respectively. Hence, for the two members we can give LY and LZ 8.5m as you have suggested. Alternatively, for the 4m member, we can assign KY and KZ of 8.5/4 and for the 6m member, we can assign 8.5/6...

The factor is applicable to all beams and columns depending on the type of connection (pinned/fixed/free) on each end. This is from Table 5.2 of old code or Table 11 of new code.

Arun

Re: steel design query

sureshprsharma — Sat, 21 May 2011 02:33:44 GMT

Dear Arun,

I once again bother you if you do not mind. Now I have partially understood your posting dated 15/5/2011. You have proposed to divide the 10 meter length of beam in two pieces of 4m and 6m each. You propose to have a uniform factor of 0.85 for the beam member. I understand from your posting that you would have assigned the factor 0.85 to the10 meter long beam even if the beam was not divided in two parts.Is my understanding correct? I further understand that the factor 0.85 would not be applied to column members. This is purely my understanding. Please clarify whether I have understood the matter correctly

Re: steel design query

sureshprsharma — Thu, 19 May 2011 22:07:23 GMT

Yours is not a case of unsupported i.e. UNB and UNT.

Re: steel design query

Arun Subramanian — Thu, 19 May 2011 21:20:04 GMT

Dear Arun & Suresh,

Should we use UNB & UNT commands in lieu of LY & LZ? Because they correspond to the unsupported length of the Top & Bottom flanges respectively..

Will it make a huge difference in the design of the member?

Kindly elaborate..

Regards,

Arun.

Re: steel design query

Arunkumar Srinivasan — Mon, 16 May 2011 11:52:08 GMT

Dear Mr. Suresh Sharma,

Yes, Ly and Lz can be used in the place of Ky and Kz and in this case, it will be easier to use...

Arun

Re: steel design query

sureshprsharma — Mon, 16 May 2011 08:32:01 GMT

Dear Arun ,

What ever you have said about Ky and Kz could not be perceived by me. I will have relook into it and try to understand

What I feel that in lieu of Ky and Kz facotos the members 1, 2 and 3 should be assigned Ly and Lz that is unsupported length, as 10 meter to all the three members. Please examine my views and give your valuable comments.

Re: steel design query

Arunkumar Srinivasan — Sun, 15 May 2011 19:43:47 GMT

STAAD uses the length of the member when calculating the allowable bending compression stress. If you use a node in the middle, you are reducing the effective length of the member. Hence, you are getting higher allowable stress which will result with a smaller pass ratio.

You have to give Ky and Kz values for the members to help STAAD identify what is the actual effective length to be taken for calculating allowable stresses. If there is a crossbeam at the middle node which provides restraint for the compression flange, then you can introduce the node and use the default Ky and Kz values. If not, then you have to enter Ky and Kz as 2 for the members with new node in center.

Calculation of Ky and Kz is Effective length factor x Actual Effective length of the beam of which the member is a part of / Length of the member...

For example, if you take a 10 m long member and cut into 4 and 6 m parts, and if you can take effective length as 0.85 x L, then

Ky and Kz for 4m part = 0.85 x 10 / 4

Ky and Kz for 6m part = 0.85 x 10 / 6

Hope that if you add these factors, your pass ratios come same in both cases...

Arun

Re: steel design query

SKLose — Sun, 15 May 2011 18:51:42 GMT

Himmatlal,

Your question is not clear.

Can you attach both the models? You have attached only one.