Steel truss - Lateral restraint forces

[RicheS]

Hi

I'm involved in a discussion with a steelwork fabricator over the correct interpretation of restraint forces to BS 5950 4.3.2.2. The Fabricator has designed to AISC 360 ASD however the specification requires design to BS 5950

We have a number of long span warren trusses spanning >50m for a large storage facility and the restraint to the truss compression chord is provided by Z section cold rolled purlins. There are plan braced bays every 5 trusses which provide restraint against the trusses buckling as a group. The roof sheeting is not assumed to act as a diaphragm (see sketch).

The force in the top chord is around 1000 kN (ASD) at mid span. BS 5950 required that 2.5% of this force (25 kN) be distributed between 'intermediate lateral restraints'. The force is cumulative with a reduction factor allowing for the number of trusses connected to each braced bay as per section 4.3.2.2.3 of BS 5950.

Is it correct to assume that the 2.5% force is simply distributed over every purlin along the entire truss span (in proportion to their spacing as per 5950) or is is distributed per truss panel? I've not been able to find much guidance on this so far.

Appreciate your feedback

EDIT Just found this useful discussion on restraint forces in parallel beams

https://www.eng-tips.com/viewthread.cfm?qid=266666

 

[HTURKAK]

I have BS 5950 2000 version.

You should read 4.3.2.2.1 and 4.3.2.2.2

In your case ,# of intermediate lateral restrains more than 3. In this case, each restrain shall be capable of resisting 1% of the compression force within the relevant span or 2.5 % divided between the intermediate restrains (which ever governs)

For your case, # of bays 4 and # of restrained truss =5. Assuming 1000 kN compression force valid for entire span,

The compression force for each restrain=1%*1000Kn = 10 Kn. (1% will govern ) And for 5 trusses (assuming the bracing is for 5 trusses as per attached dwg) Ngross= 5*10=50 Kn.
The reduction factor for 5 trusses,

Kr=SQRT (0.2+1/Nr) = SQRT (0.2+1/5)=0.632

and the design force for each restrain Nd= 0.632*50= 32 Kn.