Overlapping lintels..

[MattChan]

A blatant bump.. the BCO chap has gone AWOL so any input on if the above is acceptable would be appreciated.

 

[tony1851]

Is it going to be a 100 x 100 bearing on the lightweight block that you mean?

Without a padstone, that area of block would support an unfactored load of around 6.7Kn (approx 700kg). So if you can work out the approximate load the lintel will carry, and then halve it, if it's less than 700kg, you'd be OK without a padstone. If it's just a single door width, then it will almost certainly be OK.

But your BCO won't bother doing the figures, so will insist on a padstone; if he does, ask him precisely how big it needs to be - and why.

 

[MattChan]

Thanks Tony, its a bearing of 100 x 190mm or slightly more on the light weight block sections each side for a span of 920mm

Its got half a row of block on top of the lintel followed by the wall plate for a lean to roof with concrete tiles on. Can't see that being too heavy as it only picks up two rafter ends?

 

[MattChan]

Is it going to be a 100 x 100 bearing on the lightweight block that you mean?

Without a padstone, that area of block would support an unfactored load of around 6.7Kn (approx 700kg). So if you can work out the approximate load the lintel will carry, and then halve it, if it's less than 700kg, you'd be OK without a padstone. If it's just a single door width, then it will almost certainly be OK.

But your BCO won't bother doing the figures, so will insist on a padstone; if he does, ask him precisely how big it needs to be - and why.

Just for my interest can you run me through that calculation?

700kg x 9.80665 = 6865N

Bearing assumption was 100mm per side so 2 x 100mm x 100mm = 20000mm2

Force per mm2 = 6865N / 20000mm2 = 0.343N/mm2

For a 3.6N/mm2 block does that mean that you have to apply a safety derating factor of 10 to get from 3.6N/mm2 to 0.36N/mm2 or is my maths duff?

 

[tony1851]

Just for my interest can you run me through that calculation?

OK, keep it simple and work in N (Newtons) and mm.

When seating a lintel or beam into a wall, it produces a stress immediately under the bearing; the purpose of a padstone is to spread that load from the beam (the ‘reaction’) over a larger area of wall to reduce the stress (stress is measured in N/mm2)

The design Codes give formulae for working out what size of padstone you need; this depends partly on the reaction from the beam, and partly on the compressive strength of the bricks or blocks.

For a typical beam bearing in an existing wall, the allowable stress is given by: 1.25 x fk/3.5.

• Fk (in N/mm2) is the compressive strength of the brick or block (manufacturers give tables of the strengths of their products; Celcon is around 3.4 N/mm2; high-quality engineering brick would be 50+N/mm2).

• The figure of 3.5 is a safety factor.

So for your lightweight block, a 100 x 100 bearing would be capable of sustaining a load of 1.25 x 3.4/3.5 x 100 x 100 = 12,143N, or 12.1kN. (note; this is a ‘factored’ load – the ‘real’ load will be less, depending on what type of load the beam is carrying – If it’s a roof over a door, I’d guess the unfactored load could be up to around 8kN - but will probably be a lot less).

The implication here is that if your load is relatively low – and you have a minimum bearing length of 100 – you wouldn’t need a padstone, even on Celcon.

 

[MattChan]

Superb, thank you for taking the time to write that out Tony, very informative and educational.

My only other question is it ok for there to be a little piece of block and then a mortar join and then another block under the bearing?

The bit to the right of the red cut line will be 30 to 50mm then mortar then the rest of the 160mm bearing on another block: