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small house footings-no interior bearing walls

georgia plans exam

Silver Member
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Metro Atlanta
Footnote b to 2018 IRC Tables R403.1(1), (2) & (3) seems to require an interior bearing wall. What if the project does not require an interior bearing wall? How would one size the footings? On a side note, why do the figures at the bottom of the pages for a slab on grade not show a footing under the center bearing wall?

Thanks, GPE.
 

north star

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GPE,


IMO, ...Tables R403.1(1), (2) & (3) are examples "if" there is a
Center Load Bearing Wall...….If there is no Center Load Bearing
Wall in the project, then there is nothing to support.

It appears as though a footing is missing from the "slab on grade"
Figures in each of the Tables.


$ $ * $ $
 

georgia plans exam

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Messages
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Thanks, north star but, say I had a 22' wide house with no interior footings. Rafters and joists sized to span the entire width. The tributary load on the outside wall footings would be based on 11' of that span. If I had a 32' wide house with an interior bearing wall, the tributary span would be only 8'. So, I am not understanding the rationale behind basing the tables on an interior bearing wall. GPE.
 

classicT

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Washington State
Thanks, north star but, say I had a 22' wide house with no interior footings. Rafters and joists sized to span the entire width. The tributary load on the outside wall footings would be based on 11' of that span. If I had a 32' wide house with an interior bearing wall, the tributary span would be only 8'. So, I am not understanding the rationale behind basing the tables on an interior bearing wall. GPE.
In your example the interior load bearing wall would actually have 16' of tributary (8' from each direction).

This Table, particularly Footnote B, is so messed up that we locally got the IRC amended by the state (WA) to include a revised set of tables. Check them out at https://app.leg.wa.gov/wac/default.aspx?cite=51-51-0403&pdf=true
 

classicT

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For those that are interested, Footnote b of Tables R403.1(1) is appears to based upon the following load calculation.

32' wide house with interior load bearing wall
16' tributary load applied to interior load bearing wall
Interior load bearing wall supports 40psf of LL from floor, 20psf of roof LL, 10psf of DL from floor and roof each
Interior load bearing wall weighs 10psf (10' tall)
Total load applied is 180psf (40+20+10+10)
Load applied to footing is calculated as (trib. load) x (trib. length) + (weight of wall)
(80psf x 16') + 100 = 1380plf
1380<1500psf (assumed allowable)

Note that the exterior walls in the above example would carry 1/2 if the tributary associated with the roof and floor load, being that they carry only 8' of trib.
 
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