Another Garage Loft

I am new to this website, and found it from the other recent post about garage lofts. I have a similar question, I want to build a loft above my garage door, as I have a high ceiling. My garage is 14' wide, and I want to build the loft 7' deep. Instead of putting a hefty beam across the 14' and running joists to a ledger above the door, what if I just used 2x8/10's on like 12" centres across the 14'? I can sister a 2x4 stud to each of the ones on either side to have the joists sitting on, since that would put the weight directly onto the foundation wall below, as the front of the garage protrudes from the house, and that part of the walls is all uninsulated.
Then just some plywood on top (OSB most likely) and I have quite a bit of storage.

The other thought, is to use two 2x10's to span the front of it, with a support going to the ceiling, and then just use 7' joists with joist hangars on 16" centres to a ledger on the wall.

Thoughts?
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Welcome to the forums Luke!

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There, fixed it, try it now :-)
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Thanks

 

The deeper the support joists are, the less headroom your storage area will have. The 4" difference between 2 x 4s and 2 x 8s can mean a lot of bonks on the noggin when you're crawling around up there. Why not run 2 x 4s in the short direction, supported by joist hangers (on a ledger at the wall above the garage door and on a built-up carrier beam in front)? A lot cheaper, too.

 

Thanks, thats one thing I was thinking. What would I need to span the 14'? I was thinking either two 2x8's or two 2x10's, supported on top of vertical 2x4's on either side to carry the load to the foundation walls.

 

A carrier beam of double 2 x 8s will be significantly overstressed if loaded to the tune of 40 psf, assuming your loft is 7' deep. I come up with a bending stress of 1570 psi, which most softwoods aren't capable of carrying in bending. You could safely go with double 2 x 10s (bending stress of 962 psi). If it were mine, I'd use double 2 x 6s with a (flat) bottom 2 x 4 glued and screwed the full length, with a net bending stress of just 792 psi. A side benefit is having 2-1/2" more headroom than the 2 x 10s (deducting for the flat 2 x 4 on the bottom), in case you ever want to drive your pickup into the garage with a ladder rack or light bar on it.

 

I might be missing something here but are you saying that the stress in two 2 x 6's side by side with a 2 x 4 glued and screwed underneath would be less than the stress in two 2 x 10s side-by-side??

 

Yup. Bending stress of 962 psi for 2 x 10s vs. 792 psi for 2 x 6s with a flat 2 x 4 glued and screwed to the bottom. Numbers were computed using the conservative working stress method (F = M x c / I = M / S, where M is the applied bending moment, I is the moment of inertia and S is the section modulus). And as I stated earlier, using 2 x 6s will give you more headroom than 2 x 10s as well. Not to mention probably being a bit less expensive, although maybe a wash if the cost of glue and screws is factored in.

 

Surely the S of two 2x10s side-by-side must be greater than two 2x6s side-by-side with a 2x4 underneath? And that is assuming that the latter arrangement acts as 1double 2x8 beam, which it won't. The logic of your argument is that a deeper beam will be more highly stressed than a shallower beam, which can't be right? If you consider that two 2x8s will be overstressed, then two 2x6s with a 2x4 underneath will be even more overstessed!

 

Hold the horses! Meaning, I went back and dug out my original scratchings in the recycle bin, and found a definite error in same. I neglected to divide the computed moment of inertia by "c" (distance from neutral axis to extreme fiber in bending) to get the revised section modulus. The corrected S for a double 2 x 6 with flat 2 x 4 on the bottom is only 23.1 in. cubed, or not quite as strong as a double 2 x 8. A workable fix is to substitute a flat 2 x 6 for the flat 2 x 4, while using a triple 2 x 6 instead of just a double--works out to an acceptable bending stress of 1166 psi. Won't be quite as strong as a double 2 x 10, but will perform acceptably while still gaining some headroom. My bad for not checking my earlier calculations before posting, but that's the price I pay for scribbling at midnight. Thanx for catching my error, Tony.

FWIW--a shallow member with a horizontal bottom flange can be stronger than a deeper member having no bottom flanges. Having more cross-sectional area farther from the neutral axis makes the difference. The concept is conveyed on highway bridge girders, where the vast majority of them have large bottom flanges oriented perpendicular to the somewhat flimsy vertical webs. It's a concept that's been around for years, and has proven to be one of the most economical shapes for supporting vertical loads.