I'm enlarging an opening for a front door. I'm trying to work out the header size. Please see attached picture for reference.

Structure: As the structure of my house doesn't match what is in the code as reference points, I'm taking what I would consider the next "highest" structure (Roof, ceiling and one clear-span floor)

Ground snow Load: For my area the snow load is depicted as 55 psf. So I have rounded that up to the 70 psf column.

Building width: As shown/mentioned in the picture, the house is 20' wide. There is a little more to this however. It's a 100+ old home. It's almost as if they built two sections together but with one roof. So the one section is 20' wide by 17' deep and the other section is 20' wide by 30' deep. (the picture only shows the 20x17 section). The 20x30 section has the roof running perpendicular to the 20x17 roof. Given this, I've selected the 24' Building width.

Span: The span of the new opening is 6'. With this in mind I have two options in the 70 psf, 24' building width column: 6'7" or 6'5".

Header size: based on 6'7" I should use 3 2x12s or based on 6'5" I should use 4 2x10s.

So the question is, Am I really reading/understanding this correctly? 3 2x12s 'feels' like overkill for a 6' span.

(Just a side note, currently there is no header over the 3' opening!)

I've attached the table section of the ICC IRC 2021 code for reference.

• Wow…50 psf snow load is a lot. Do you live at the top of Mt. Everest? Jul 4, 2021 at 18:16
• We have a 50 in Michigan, Our highest point is Mount Arvon at 1,979 ft.
– Gil
Jul 4, 2021 at 19:27
• @LeeSam 50psf is for when it snows and then rains on the snow. Typical for wet winters, where it can rain before the snow is melted. Jul 4, 2021 at 21:29

You didn’t provide the footnotes to the charts, so let’s do this longhand.

Therefore, the total design load is 85 psf plus 55 psf = 140 psf plus wall load above header.

Wall load = 5’ x 15 psf = 75 plf

Therefore, the total design load is 140 psf x 12’ (half the building load goes to the other wall in a 24’ wide building) = 1680 plf plus 75 plf = 1755 plf on the header.

Depending on the species and grade, a 2x10 will support about 480 plf for a 6’ span and a 2x12 will support about 680 plf for a 6’ span.

Therefore, it requires 4 - 2x10’s or 3 - 2x12’s…just as the table suggests.

The big question is the suggested snow load of 70 psf. How accurate is that?

• Thanks @Lee Sam! Great detail, very much appreciated! With regards to the snow load, in my area according to the a couple of sites it is 2.8 kPa. (I am just east of Georgian Bay, so we get wet lake snow. ) If I have done my math correctly, 2.8 kPa translates to almost 55 psf. Jul 5, 2021 at 13:43
• Thanks again @lee Sam. The 4 2x10s will be 6" wide. The studs are 4" wide (its an old house with real 2x4s). Given this, there would be a 2" overhang of the header where the studs would not rest on. I can accommodate the 2"s on the outside but I'm unsure if that is right thing to do. Any suggestions? Jul 5, 2021 at 20:38
• If you can use 55 psf snow load instead of 70 psf, I’ll recalculate the beam size. At 15 psf less, the roof load is 70 psf + 55 psf = 125 psf x 12’ = 1500 plf + 75 plf = 1575 plf. Therefore, about 3 - 2x10’s can support the load for a 6’ header. (It’s slightly overstressed, but the Code allows for such loading.) If this doesn’t work, you could use an LVL.) Jul 5, 2021 at 21:03