Spans for ceiling

[Bob M]

Hello, I'm building a fully independent suspended acoustic ceiling in two rooms of an apartment. The frame will be completely freestanding — sitting on the masonry walls only via a timber wall plate, with no connection whatsoever to the existing ceiling above.

Room 1: 3.2m span
Room 2: 4.18m span

Ceiling load (dead load only, no imposed load):

• 2 layers 15mm acoustic plasterboard: ~29 kg/m² (0.28 kN/m²)
• Optional Tecsound 50: +5 kg/m² (0.33 kN/m² total)
• Plus self weight of joists

A soundproofing specialist has told me 100-120mm (4-5 inch) deep timber will be sufficient. I've seen online that span tables suggest I need 145mm for Room 1 and significantly more for Room 2, but the specialist argues the tables are overly conservative for this application as they account for imposed loads like storage in loft spaces that simply don't apply here.

Also, in this video (

), another specialist says that for a 4x4m room they would use nothing less than a 5x2inch timber.
My questions:

1. Is 100mm or 120mm timber realistic for these spans given this is dead load only with no imposed load?
2. What size and spacing would you recommend?
3. Does adding Tecsound meaningfully change the timber required?
4. Would 300mm centres meaningfully improve the permissible span, and is there a calculation method or reference for non-standard spacings?

Thank you so much in advance!!!

 

[stevie888]

Timber is funny stuff: it delects a very long way before it breaks and all the TRADA tables are based on permissable deflections. It is highly likely that even 100 timbers won't break over that span but the ceiling won't be very flat.

Youve got 2 specialists giving you advice which doesn't actually wildly disagree: 4/5 vs 5 is hardly majorly different

If you want to calculate then (either manually or) using a program such as Superbeam you design a beam to carry a 1m strip of your ceiling so all your loads in kN/sqm are the same numeric value as distributed loads in kN/m. Set yourself a beam depth and your answer is a width of timber required for that 1m width.

Say it comes out at 180mm: icf you put nominal 50s in at 250crs you'll have 4x47 (true size) or 188mm