I-beam size 30' span

[dieselcrawler]

We are thinking of adding on to our house, 24' deep, 28 to 30' wide. The lower level will be garage space, and it would be nice to not have post in between the parking spaces supporting a beam for the floor above. The beam would be spanning the 28 to 30' width, so I can use 12' floor joist for the 24' depth. The ceiling hight in the garage area will be 10 to 12' tall, so the hight of the I-beam wouldn't matter, as I can easily keep a 9' clear space below the beam. The floor above will be bedrooms and bathroom area, I do not remember the loads per square foot for calculations, been 20 years since I had to think about it.

Any feedback welcome, just in the brainstorming stage right now, starting some sketches, to get materials estimated, and cost figured.

 

[schmism]

40 lbs/sf live load for floors (40psf LL) is typical for residential construction.

 

[CurlyDave]

I would design this with the beam in the 24' dimension.

Most span tables will allow much more than 15' for 2" x 12" joists of almost any species commonly used for construction.

A 24' beam can probably be a GluLam using standard joist hangers for the joists.

24' x 30' = 720 sq ft. live load of 40 psf = 28800 pounds. 1/2 is supported by beam in center = 14400 lb

14400/24 = 600 pounds/lineal foot.

Look at the table here: http://www.aitc-glulam.org/pdf/Capacity/DF_27.PDF and you would need either a 5-1/8" x 18" or a 6-3/4" x 16-1/2" Doug fir/Larch GluLam.

This is not a particularly large beam and I would increase the size to 6-6/4" x 18" or even 19-1/2" to give a substantial increase in floor stiffness at a very modest cost increase.

P.S. when you have this designed consider adding a hard point in one bay of the garage suitable for a chain hoist or even a come-along to use as an engine hoist. I suggested this to my neighbor about 20 years ago, and he put one in. When his then 8 year old son entered his teen age years they must have rebuilt 3 or 4 engines together, always using the hard point to remove and replace them.

 

[the old grind]

Brainstorming? (insomnia) My home's main floor is on 18" deep 2x4 'W' trusses 24" OC and interior walls are net 27' apart above where they span. There is no beam, and there's a raceway for the HVAC plenum down the truss centers. Family room below is 18' x 27' with no posts or beams. ('residential height' ~ 7' 6" to suspended ceiling. (pics on request .. )

The steel beam proposed I'd guess to be all 12" or more unless say two set spaced for 8' joists above or other combo. (It/they could be between vs under joists if you add a lintel(s) to the bottom(s). Door header need not be below them but needs considered depending on which wall gets doors and their sizes/clearances. Ceiling might be higher than the bottom of a beam under joists (assuming 12" & 12"). Do check with an engineer, familiar pal/dealer about options/possibilities for such trusses, beams, lintels, etc (16" OC surely spec'd) vs simple joists & beam and compare costs of all the bits each thing considered.

Ease of pluming/wiring facilities above/thru trusses (BTDT) is nice. Much to consider. Door header might could be tripled joists if on the short wall vs welded up or something. Doors on the long wall? 'W' trusses are hung by the top chord, so a header under their tails would o'lap nearly their full depth/height. (my garage's ceiling and door header are trussed.)

Your building inspector/dept might refer you to someone for advice to their 'er liking (sure to green tag first insp). There are ways ... and I apologize if I added confusion to a process whose method has been pretty much decided.

Oh, and my guess on the beam 'depth' was based on a 28' wide x 82' modular home setup. Going from 4" x 8" to 4" x 12" steel to span the four main 'rails' (~27') was approved to eliminate all posts. (Going from 4" to 6" wide beam would mean more of its own weight/load to carry, to install, and +$$). Garage, entry from narrow end, is 27' x 30' and adjacent family room is 27' x 30.5'. btw, For o'all ceiling height one 'cross-beam' in each ceiling (4" x 4" x 28' showing) is cosmetically wrapped but not obvious/unsightly. tog

 

[dieselcrawler]

I have to span the 28' distance because I cannot put one in the center to span the 24' distance, it would be in the middle of one of the garage bays. If I went that route, I would have to use 2 beams 24' long,

I also doubt any major automotive repairs would take place here, as I have a work Bay in the barn already set up with beam trolley and 2 chain hoist. Not that I wouldn't mind, this garage would be bigger and nicer and warm and... and the wife won't let me! This is "her" parking area, I've already "junked up the barn enough".

Also, we are out in the country, there is no inspection to worry about, but I, like my father, want to do things right and tend to lean towards overkill... if it calls for a 10" beam, I will likely get a 12" beam.

 

[LD1]

Assuming 50PSF live + dead load and 30' span supporting 1/2 of the 24' width.....is 18k lbs

Thats 600 plf along the 30' span of the beam. For floors, deflection is liked to be kept L/360 or better. So allowable deflection is 1"

W10x68, W12x50, W14x38, W16x31 all meet the criteria. Those would be what I would call minimum for the size class. If you want to go heavier, Something like a W14x43 or W14x50 instead of the W14x38.

But, those already have some SF built in, since it is unlikely that you will ever have 50# on every square foot of the above floor. But thats what you should design for. Lots of areas may have WAY more than 50PSF. Like a bathroom with a tub full of water and you in it, or a gun safe, etc.

I am unsure why you said you cannot run the beam the 24' direction? What is going to be in the middle of the garage? The beam will clear span to the outside walls...

The 24' direction and supporting 15' of the long direction would still be 18k (still supporting exactaly half the load), but a shorter beam will now have 750plf on it. And 24' length, you would want to keep deflection under 0.8" now.

Beams for that would be: W10x45, W12x30, W14x26

A 24' long W12x30 is gonna 720#. A 30' long W12x50 is gonna be double that. So consider the weights and the equipment you have

 

[LD1]

I understand not being able to run the 24' direction now.....

Have you considered the 2-beam approach? A beam at Wall at 0', beam at 10', bean at 20', wall at 30'.

Would shorten the joist spans and may save some $$$ there getting by with 2x8 or 2x10's, or spacing different or a combination of things.

Each beam would catty 24' of floor and 5' to each side (10' total). So 240sq ft x 50psf is 12k lbs total, and 500plf

W8x48, W10x30, W12x22.

Cost difference on a pair of lighter beams @ 24' isnt gonna be much different than a single heavier 30' beam.

I'd consider the rest of the design, floor joists sizing/span, etc before I arrived at a final conclusion.

 

[dieselcrawler]

The 28 to 30' wall will have 2 garage doors and a man door, with center of the span ending up about 1/3 in from one edge of one of the garage doors. Not interested in having a header over the door support a 24' beam.

With the existing grade, existing floor levels, the garage floor and ceiling will have plenty of room to allow a beam of adequate size and still not hinder head room.

The floor over the garage will match up to the existing 2nd floor of our house, for smooth transition from current hallway to new section. First floor has 9' ceilings, and is at least 3 feet above current grade outside, and that will be even lower once leveled as needed for construction. I'm ok with high ceilings in the garage, I can utilize the space for storeage, by hanging large shelves from the ceiling above, hence my 'overkill' building methods.

Beams quoted in post above are an inch dimension x weight per foot, right? What is the corresponding inch dimension, that is what would be most helpful to me at this point of my design process.

Example: 'to span 30' as described above, you need a beam that's xx" tall, yy" wide and weighs zz lbs per foot.'

Ballpark close enough, I'm likely to oversize a bit anyway, and double check that everything is good once I get further along. It could be another year before I even start this project.

 

[LD1]

The sizes I referenced are the standard sizing of the beam. Thats what you would specify when calling to order one.

Use this chart Steel I Beam, Wide Flange Beam, S, Jr., WF, details, sizes, dimensions, beams

IE: the W12x50 I gave you is W= wide flange 12 = 12" depth (height), and 50# per foot is what it weights. That coorsponds to a beam that is (nominal size) 12" tall and 8" wide.

But be careful. That dont mean every beam that is 12" high and 8" wide will work. IE: a W12x40 is ALSO 12x8

And using the 30' span and 600plf along the beam, the

W12x50 will deflect 0.925" (under the 1" limit of L/360)
W12x40 will deflect 1.18" (too much)

 

[LD1]

Also, if you plan on using used beams sourced from somewhere, You can use that chart, a tape, and a pair of calipers to know exactaly what you are looking at.

The W10x68 is a 10" x 10" beam. And there are 8 beams with those nominal dimensions, 3 of them being inadequate
The W12x50 is a 12" x 8" beam, 3 in that size. The other two are inadequate
The W14x38 is a 14" x 6-3/4" beam, 3 of those and the other two are inadequate.
The W16x31 is a 16" x 5-1/2" beam. Only 2 of those and the other is smaller as well.