Brownstone foundation wall

[JONNH]

Hi All,
We are working on an existing brownstone and are adding some additional backfill to the exterior of the front foundation wall, in turn increasing the soil loads on the wall. In going through the calc I am finding the wall is not close to working, even for the existing condition.

Here is an example:
Say you have a 10ft tall foundation wall, 16" thick. This is a 3-story brownstone, so lets say the exterior wall above is 12" brick, and when accounting for windows has an average weight of 100psf. The floor framing typically runs parallel to the front and back walls, so no additional floor dead load. This gives a dead load of 30ftx100psf=3,000 plf.

Assuming a 16" thick foundation wall and that the dead load acts through the centroid of the wall, the overturning resistance is 3,000lbs x 16"/2=24,000 lb-in=2,000 lb-ft.

Just looking at the soil loading with an equivalent fluid pressure of 60pcf, we have a moment of Msoil=.128*(60pcf x 10ft x 10ft/2)*10 ft=3,840 lb-ft.

With out any factors of safety applied or including surcharge and seismic loading, we have a resisting moment of 2,000 lb-ft vs 3,840 lb-ft applied.

This would appear to be a very common wall in these parts, am I missing something here??

Thank you.

 

[oldestguy]

I'd consider the wall as a vertical beam with one support at the foundation and the upper support as the first floor, even though the joists are parallel to the wall, but using the flooring on the joists for the reaction transfecting that load to the joists and flooring combination. Practically have you ever seen a sideways movement of joists parallel to the wall? I have not.

 

[JONNH]

Thanks for the response. I am considering it pinned top and bottom. The only reason I mentioned the joist span was to point out that no additional dead load is added by the floor.

The moment equation in my original post is the max moment for a pinned=pinned beam with a triangular loading.

Thanks.

 

[oldestguy]

OK on the pin-pin setup. Next the loads. A common number to use is 1/3 the vertical pressure, but the lower part of the triangle is cut down so the rough load diagram is near parabolic. With even small percentage of fines (silt and clay) the cohesion comes in to sometimes reduce the horizontal pressure to zero. Look up critical height. H=2c/sigma H= height of tension crack. c= unit cohesion . sigma = unit weight of soi.l English units.

 

[fattdad]

when the wall is restrained at the top and bottom, 1/3rd of the vertical earth pressure is too low. I'd stick with the 60 pcf.

Not a structural engineer; however.

f-d

ípapß gordo ainÆt no madre flaca!