How much wall does a lintel need?

[dot.com]

The back of my house is effectively a single storey (house is built in to a hillside) and the wall is a block/brick cavity wall supporting a roof of interlocking (concrete?) tiles. It's a 1990's Wimpey construction. In this wall is a 1200mm wide window and a 1800mm french door, both have some form of cavity wall steel lintels.

The outer leaf of brick doesn't climb past the lintel, so the soffit board just rests on top of the lintel/bricks. Inside there is about 300mm of block above the lintel, before it meets the wall plate.

I'm planning to make the window much wider, around 2800mm, and make the doorway narrower to around 1000mm. I will need to build up the brick/block to make the doorway narrower, and that 'new' bit of wall will become the bearing for both the window and door lintels.

The resulting 'middle' bit of wall between the window and the door will be about 737mm wide, and provides a 150mm bearing on each side, one for the door and one for the window. How can I work out if this section of wall can support the loads?

The roof span is 9400 and 35 degrees.

If I assume worst case the block is 2.9N/mm2 then I calculate 737 x 100 x 2.9 = 213,730N but I have no idea if that is representative of its capacity to support lintels.

Any advice very much appreciated!

-phil

 

[noseall]

The back of my house is effectively a single storey (house is built in to a hillside) and the wall is a block/brick cavity wall supporting a roof of interlocking (concrete?) tiles. It's a 1990's Wimpey construction. In this wall is a 1200mm wide window and a 1800mm french door, both have some form of cavity wall steel lintels.

The outer leaf of brick doesn't climb past the lintel, so the soffit board just rests on top of the lintel/bricks. Inside there is about 300mm of block above the lintel, before it meets the wall plate.

I'm planning to make the window much wider, around 2800mm, and make the doorway narrower to around 1000mm. I will need to build up the brick/block to make the doorway narrower, and that 'new' bit of wall will become the bearing for both the window and door lintels.

The resulting 'middle' bit of wall between the window and the door will be about 737mm wide, and provides a 150mm bearing on each side, one for the door and one for the window. How can I work out if this section of wall can support the loads?

The roof span is 9400 and 35 degrees.

If I assume worst case the block is 2.9N/mm2 then I calculate 737 x 100 x 2.9 = 213,730N but I have no idea if that is representative of its capacity to support lintels.

Any advice very much appreciated!

-philView attachment 346982View attachment 346981

3.8m span, with direct loading above? Steel beam territory I'm guessing. Why are those French doors showing a faux header? Are they concealing summat?

 

[dot.com]

The doors just have a piece of timber above them, not sure why they didn't install doors tall enough to meet the lintel.
Yes definitely steel, Catnic are suggesting one of these
https://catnic.com/assets/datasheets/lintels/cn81-c.pdf
But I'm asking if I can use one of these instead

TX90/100 Extra Heavy Duty Steel Lintel | Lintels | Catnic

TX90/100 Steel lintel with unique thermal break for use in cavity wall construction, designed to support typical masonry, timber floor and roof loads…

catnic.com

which has a slightly higher SWL. Their thinking is the outer leaf toe will be exposed and have the soffit resting on top of it if I use a normal cavity wall lintel (as it does today), so thy're suggesting I use the more boxy version. I don't want to use that because it will require me removing 219mm more block, meaning I'll only have about 90mm of block between the lintel and the wall plate, and I really don't want to remove more block than I need to. It would be difficult to cut that close to the ceiling, and hard to support it all in the process, and I really don't want to have to peel back the roof to access/expose the wall plate etc. I was hoping for a simpler remove/replace operation!

 

[dot.com]

3.8m span, with direct loading above? Steel beam territory I'm guessing. Why are those French doors showing a faux header? Are they concealing summat?

I was planning on 2 separate lintels, one for the window and another for the door

 

[dot.com]

 

[dot.com]

Can anyone comment? How much wall is not enough?! I assume there is a reasonably simple way to calculate how much wall is needed to support a given load. Am I worrying over nothing?

 

[noseall]

Can anyone comment? How much wall is not enough?! I assume there is a reasonably simple way to calculate how much wall is needed to support a given load. Am I worrying over nothing?

450mm for a pair of lintels landing on masonry although technically, it could be less.

 

[Dereekoo]

450mm for a pair of lintels landing on masonry although technically, it could be less.

How can you say that without knowing what the actual load is ?

 

[noseall]

How can you say that without knowing what the actual load is ?

True.
However, looking at the OP''s scenario and knowing that proprietary lintels require 150mm end bearing, I gave it my best shot.

If its any comfort, we've loaded steel beams onto less.

 

[dot.com]

Thanks for the comments; I decided to engage an SE to put my mind at rest

 

[Dereekoo]

If I assume worst case the block is 2.9N/mm2 then I calculate 737 x 100 x 2.9 = 213,730N but I have no idea if that is representative of its capacity to support lintels.

Look at worst case scenario of load acting from half of 2.8m lintel on a bearing area of 150 x 100 mm

Assume front half of roof load is transferred to front support walls (unlikely to be the case because I assume you have some internal walls in the bungalow also taking load). Taking a load for the roof and ceiling and snow load of 2kN/m2 (generous) will give a load acting on the front wall of 2 x 9.4/2 =9.4kN/m run of wall. Multiply this figure by a standard enhanced loading factor of 1.5 gives 14.1kN/m run.
Load acting on end of lintel due to roof loading = 2.8/2 x 14.1 = 19.74kN
Bearing pressure = (19.74 x 1000)/(150 x 100) = 1.316 N/mm2 >2.9N/mm2

As stated this pressure is a worst case scenario and will quickly reduce as it dissipates downwards at 45 degrees. I assume your SE will quickly put your put your mind and wallet to rest