CLIFF NOTES: help me calculate a beam for my uncles house.
well guys, I am behind on classes for my architecture major, and my uncle asked me to help him figure out what size beam he needs for his house that he is remodeling. I had a few of my friends try and help me out and all we could figure out is he needs a 9" beam based on pre-fab beams. he needs it to be under 5" and he knows a place to have one custom made, so he needs to know how thick the “webbing” would have to be on the beam to be able to handle the load of the floor above.
Specs:
span of beam would be 16ft, no post in the middle to help with flex…just at the ends. it would be directly over the beam that is in the basement that is supporting the first floor. and the posts will be on the existing foundation walls, burried within the interior walls. (brick and mortar footings)
the beam is to support half of the house’s second floor. the ceilings above the second floor are cathedral, and an attic is only over half of that second floor.
i calculated the square footage of that half of the house plus the attic(728sq.ft.), and figured on the standard of 30# per square foot for the weight rating needed.
i came up with 22,000#s, or 1,375# per square foot of a 16ft beam.
so a friend of mine that took structures 1 and 2 tried to help me out, and all he could do was run the numbers and put it through a table of available standard beams. the value he was running through was a “K” of 1375 multiplied by the length of the beam (16ft) and divided it by 2 to get a number of 4400K#/ft. he also got some “S-value” of 42.24 cu.in.
running all the numbers yielded a 9" beam, which he cant use. i was wondering if he could use two 5" beams side by side, but im not sure he wants to do that. i realize this is a huge thing to read but if anyone can help me out ill throw ya $20 for your troubles.
Future cliff notes: Paid $20 to an undergrad architecture student from a local car forum for advice on a critical load bearing portion of uncles house and second floor collapses when aunt takes her first bath upstairs.
I am not nor never was an architecture student, but there is a lot you are leaving out. Are all the areas of the house involved here contained within the original structure, or does this include some type of addition? If it is solely within an existing structure, is the attic supported only by the outer walls, or does it utilize interior load bearing walls that rest on the second floor (or in the case of a remodel, walls that may have continued through the first floor).
I think to get any kind of good answer on this someone would need to look at it in person, and that person should be a professional.
the remodeling has so far kept all existing walls. the home is a previous 1-room school house of the very old days, the owner before my uncle converted it into a two story home, poorly. the area above the beam is 2bedrooms, with the attic above one of the rooms where the load is on two of the outer original walls on the corner, and on the middle wall directly over where the beam would go separating the two rooms. the way the current first floor is oriented is the same as the two bedrooms above, where downstairs is a den/office and a living room space. my uncle wants to eliminate this central load-bearing wall to make a nice open larger living room space and replace the wall with a beam that would support the existing upstairs.
EDIT: i had tried emialing the teacher of the class, but she never replied. im assuming because she has no idea who i am because i was never enrolled in her class and probably got cut by the email filter.
EDIT2: if anyone is serious about helping me I would be more than happy to drive them to the place on a weekend and buy dinner in addition to the $20. however, it is in “the alabama swamps” aka the boonies past lancaster speedway
i could solve this for you in like 2 min if i still had my steel book from school. Using the values your friend calc’d, assuming he’s doing it correctly, you can look through the tables in that book to find a 5" beam that should fit what you need, if its not in that book it will have to be a custom beam which is going to be big $$$$. Just make sure your friend knows what he’s doing with those calcs, i’d bump up the original weight to an even 1500 #/sqft just for a little added safety. And are you 100% sure that beam is only going to take load from half the house? This isn’t something you’re going to want to guess on unless you’re willing to risk your uncle’s life.
i came up with 22,000#s, or 1,375# per square foot of a 16ft beam.
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You mean pounds per linear foot. Hopefully that is what “K” is supposed to be.
This is a great exercise for you. Go through it all, solve it, grab a mechE’s Mechanics of Solids book if you need to calculate steel beam dimensions, but ultimately don’t tell your uncle what to do. Get a professional evaluation and compare it to your own, and figure out the discrepancies.
Imagine if you told him what to do, he did it, and you were wrong. At best there’s a family feud. At worst there’s a lawsuit.
i take all this constructive critisizm to heart. he doesnt want to hire a proffesional, and he knows a guy that will get him a custom beam mad cheap so he doesnt care about cost. he already has been warned im not responsible if im wrong but i would give it a wack for him. the reason he came to me is that beam manufacturer’s refuse to give a rating on the beam because of lawsuits like you guys are talking about. they say only an architect can rate a beam and all they are responsable for is the order.
Gotcha. If I were you I’d try and get some one on one time with an architect professor, which I think you’re already working on. A good one should be pretty responsive to you expressing excitement to have the chance to actually apply what you’re learning to a real-world problem.
just let you know that you have to also know your material…not just choose the specific size beam. Also, take into account other things, beyond “will the beam hold”.
You’ll need to know your allowable stress value. For A36 Steel, you have a Yield (failure) of Fy = 36ksi grade material, 992 steel, you have a Fy = 50ksi. Respectivly allowable stress of 24ksi and 33ksi.
According to you value of 1375# per ft… in structural steel applications that’s 1.375 (Kip-ft) or the variable Wc.
Example beam that would “work” assuming your 1.375 Kip-ft value.
Total Allowable Uniform load for this beam (16ft span) = 2.375 kips (which exceeds your value)
End Reaction = 1.1875 kips
Deflection = 1.04" deflection in the beam
So what i am showing you…is even though a smaller beam can support the weight and technically not Fail, …you will have deflection to worry about. Especially if you’re looking to span a distance with as little material as possible within a residence.
My disclaimer… those are just the beam calculations and don’t utilize any of your calculations.
^ ok, so now what do i do with this? it looks like a different language to me. all im looking for is what beam material and webbing thickness in 5" tall will span 16ft with minimal deflection and handling the given load of 1375lb per linear foot.
yes i did, and my friend figured on a w9x(something) being just barely adequate. is it possible to put (2) w5x(something)'s side by side to equal the same value as the 10" beam?
can these calculations be manipulated to change the thickness of the webbing on the beam? i know that on normal beams the webbing is a ratio to the height of the beam, but my uncle can get any thickness in any hieght he wants through a friend of his. what about a “boxed-in” i-beam? like weld sheet metal on the sides to cap it all off?