What would be the minimum solid-wood beam dimension to support a 10' span supporting 10' floor joists only for a non-bearing wall area. (ie: we are removing interior wall between a kitchen and dining room...a bedroom floor only is located above kitchen and dining room). There is no weight bearing wall located above the span, only a bedroom floor built upon (approx.) 12' floor joists.

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    Commented May 2, 2019 at 7:19

1 Answer 1


The Code requires a Live Load (people, furniture, etc.) of 30 lbs. per square foot plus all Dead Loads (joists, subfloor, underlayment, carpet, etc.) Dead load is usually about 10 lbs. per square foot.

So, the Total Load is a minimum of 40 lbs. per square foot.

If your new wood beam has joists that are 10’ long resting on it, you’d have a load of: 10’/2 x 40 psf = 200 lbs. per linear foot. (Half the load will rest on the beam and half will transfer to a wall or other support.)

However, if the new beam has 10’ long joists resting on it from both directions, then you’d have a load of: 10’ x 40 psf = 400 plf.

Therefore, if joists are just spanning one way onto the beam, I’d use a 4x6 spanning 10’ (allowable load is 211 plf), depending on the species and grade. However, if joists are resting on the beam from both directions, I’d use a 4x8 spanning 10’ (allowable load is 400 plf) depending on the species and grade. (I’ve used “select structural” grade for both.)

For both beams, I’d use a 4x4 post and connect it to the beam with a steel connector, like this from Simpson:


Also, I’d make sure the post sits on a footing and is connected to it with a Simpson connector like this one:


I don’t understand the 12’ span info added at the end.

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