hello everyone
i am design the steel building i staad pro . now if the beam is continuous say 10 m and i am having the node at the ends. the say i'm getting the beam section as pass with iswb 450 and if i am inserting the node in between the beam say one at mid point the the staad gives beam as pass with value less than the above written value. say ismb 400 what is the reason. i have check all the parameters and they are ok . but design is not sufficient as per requirement.
Himmatlal,
Your question is not clear.
Can you attach both the models? You have attached only one.
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
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.