Hi. What size concrete lintel would be required to safely support the brickwork in this chimney. The width is 800mm. So, I understand that 1000mm lintel is required. What I can't quite get my head around is the dimensions of Width and height. Would a 4" X 3" lintel suffice. Or 6" X 3". I've looked at how to calculate lintel size requirements but struggling to find info. Many thanks.

It depends on what you want to do. If you remove the plaster just below the picture frame to the current opening you will probably find an original lintel or arch. If you want to fit something like a wood burner you may have to do a lot more work to open up the chimney to accept a pipe which can be quite tricky. If you want it purely as an open space you will need to fit an inner shroud to stop debris falling down and the heat going up

Load acting on lintel is basically the weight of a 45 degree triangular wedge of brickwork acting from above the opening, Assuming a 800mm opening and weight of 100mm thick brickwork being 200kg/m2 the load on lintel= 0.8 x 0.4 x 0.5 x 200 =32 kg. Give this a factor of safety of 3 then lintel is required to take a load of 96kg . A standard 65 x100 lintel spanning 800mm can take 1000kg, well above your requirements

Hi all. Thank you for your replies. Severntrent. Thank you very much for your detailed response. It's great. Would you please explain the calculation. I.e 0.8x 0.4 and so forth. What do each of the numbers represent? I'm not querying your calculations, I'm very interested in how it is done. Thanks again.

Area of the triangle above opening = width x half the height = 0.8 x 0.4 x 0.5 = 0.16m2. Density of bricks say 2000kg /m3, hence 200kg /m2 for 100mm thick brickwork Total load of 0.16m2 of brickwork = 200 x 0.16 = 32 kg Multiply by at FOS, originally used 3 but 1.5 would be more normal which gives a design load of 48kg (0.48kN)

Hi. Would half the height not be 0.2 then. Your picture indicates 400 height. Also what does the 0.5 correspond to in the 0.8 X 0.4 X 0.5 Ta

You have a structural engineer on this forum giving his time and experience for nowt and you seem like you’re arguing the toss. What gives?

Ok, I understand. I think. Thank you JustPhil Jord86, thank you also for your input. I can understand that you thought I may have been arguing. However, JustPhil was correct. I was simply trying to understand. And if I was correct that the equation was incorrect (I wasn't right, ha) It may have proved I was getting the hang of it. Thanks again all