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ANSWERED on Thu 26 Apr 2007 - 2:26 pm UTC by redhoss

Question: sizing steel I-beam

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Tue 24 Apr 2007 - 10:13 pm UTC



I want to build garage with a 24' span with attic above. What size steel I-beam
do I need. The floor above supported by the steel I-beam will have 2x12
joists and 3/4 inch OSB plywood plus my junk in storage.



Thu 26 Apr 2007 - 3:13 am UTC

Uclue Researcher Request for clarification


Andy, I think I can help, but I need some additional info from you. I understand that the beam spans 24'. What is the garage dimension in the other direction. I need this to calculate the beam loading. Also, is the "junk" you refer to unusually heavy. A live load of 20 psf is commonly used for attic storage areas. If your junk is heavier that what you would normally store in an attic, we might want to use 40 psf which is the number used for living areas. I assume that this beam carries none of the roof load and the trusses rest on the walls. 


Thu 26 Apr 2007 - 11:34 am UTC



The garage dimensions are 24' x 24' with 2x6 walls 9 feet tall. I want to place the 2x12 joists on top of the I-beam if headroom below allows, if not I will need joist hangers from Strongtie. I don't plan on storing anything unusually heavy but what if, maybe I should go with 40 psf to be sure.
I have been wanting to add a trolley (rides on the bottom plate of the I-beam) here is a link to cut and paste. ( to the I-beam just for fun. I probably wont ever use it but if I did I would probably need to temporarily support the I-beam with a post in the middle of the span. I thank you for your time and consideration,


Thu 26 Apr 2007 - 2:26 pm UTC

Uclue Researcher Answer


In this case deflection is the determining factor. The formula used is:

I = 5wl^4/384ED

w = (live load + dead load) x 12 = 50 x 12 = 600 #/ft
l = 24 ft
E (modulus of elasticity) = 30,000,000 #/sq in
D (allowable deflection = 24 x 12 / 360 = 0.8 in
I (moment of inertia)

I = {(5 x 600 x 331,776) / (384 x 30,000,000 x 0.8)} x 1728

NOTE: 1728 is a conversion factor to get the units correct in^3/ft^3

I = 187 in^4

I checked this with a beam design program and got the same result.

A good choice would be a W12x27 with I = 204. This is a 12 inch deep beam that weighs 27 #/ft. I don't know how your headroom will work out, but you could go with a heavier beam that is 8 or 10 inches deep.

Please ask for a clarification if there is any of this you don't understand or you want to look at a beam of less depth.

Good luck with your project, Redhoss


Fri 27 Apr 2007 - 3:15 am UTC

Request for clarification


I called 2 I-beam dealers and both said only a 26lb and 30lb is available. I requested the 30lb because it is heavier, Is the 30lb a better choice than the 26lb? Also, will the w12-30 support 40psf dead load?


Fri 27 Apr 2007 - 2:15 pm UTC

Uclue Researcher Answer clarification


I guess my 1977 version of the "Manual of Steel Construction" might be a little out of date (just like me). Here is a current list of 12 inch wide flange beams:

The W12x26 shows a depth of 12.22 inches and an I = 204.
The W12x30 shows a depth of 12.34 inches and an I = 238.

So, the W12x30 is slightly stronger in bending and slightly heavier than the W12x26. Either would be just fine for your project.


Fri 27 Apr 2007 - 9:16 pm UTC

Accepted and rated


Redhoss was quick and professional with his answer. I will use Uclue in the future thanks to how well this worked out.


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