What criteria are used for sizing a beam
I have a W8x18 beam spanning 31′ over 2 equally spaced columns (continous beam), and I’m investigating putting one column in the center and removing the two existing columns to make the finished basement more friendly to the pool table and other uses.
The beam supports a floor above measuring 22′ x 31′. The floor in turn supports some partition walls, but those walls only support ceiling joists on one side of the beam (long bearing wall is over beam), while the ceiling on the other side is supported by trusses running parallel with the steel beam, spanning the full 31′. The roof doesn’t appear to be supported by any part of this system, though.
Bending stresses don’t appear to be a problem, even assuming A36, but deflection is going to be much more than before, even if modelled as a continous member. What I don’t know is what criteria are used in sizing a beam. How much deflection is desireable (existing is waaay under L/360 using 40+10 psf), and how would you reinforce it if needed? I’d rather not weld cover plates (top and bottom), as it’s in service in a wooden structure, but would prefer to sister wood or steel against the web (two C6x13’s will increase I from 61.9 to 96.7 in^4), if possible, just to increase the stiffness of the two new spans.
I’ll seal the drawings myself, but I don’t work in building construction (I design very large machinery), and therefore don’t know what the codes and conventions are for sizing such a beam. Any suggestions would be appreciated, as always.
Be seeing you…
Replies
On a beam that carries floor load, I wouldn't use anything less than L/480.
Personally, I don't like the idea of a W8X18 beam spanning over 15'. Seems pretty thin to me.
Reinforcing it would be difficult, I suspect. I have no exerience in that area. It might be best to get some advice from an engineer who has some experience in doing that. It may be cheaper just to replace it if you get that far into it.
That does seem to be a stretch. My archy had me use a W10x22 w/ 4" columns every 11'6"
On a beam that carries floor load, I wouldn't use anything less than L/480.
Why would you need the beam to be designed with more stringent criteria than the floor it supports?
Jon Blakemore RappahannockINC.com Fredericksburg, VA
"Why would you need the beam to be designed with more stringent criteria than the floor it supports?"
I wouldn't design a floor to less than L/480 either.
A beam can contribute to the vibration of a floor system. So if a floor member is CLOSE to having vibrations, an undersized beam can change the vibration enough that the floor would feel unacceptable
Little Miss Muffet sat on her tuffet, and what a big tuffet she had.
So if you're feeling insecure, just stand next to her, then you won't feel quite so bad.
TKanzer:
I have to agree with Boss Hog & Soul Train. That's pushing the envelope pretty hard. Even if the beam would make the span I think that it would feel like a trampoline as you walked across the floor upstairs. If you're really serious about doing it, I'd be inclined to get an engineer involved and get him to spec the reinforcement that would be needed.
BILL
I'm not quie picturing all the loads, but it sounds like this is already pushing the limits for an 8x18 to me just goiong by the seat of my pants.
But the last time i checked, I don't get many skid marks in that seat. Removing one of those columns would definitely change things there.
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For comparison purposes I have a W8x28 spanning 23' and supporting a 23' wide floor area (11.5' each side of the beam). I'm not looking at the calc but it was slightly better than L/480.
The L/ value is only for maximum deflection (e.g., to prevent plaster damage). The problem would not be the maximum deflection, but how easily the beam bounces from minimum dead load (including furniture) to maximum deflection when one or two people enter the space. Like BossHog and the others posted, it's the bounce effect....that's not a mistake, it's rustic
"Like BossHog and the others posted, it's the bounce effect."
That's what I'm looking for. By sandwiching a pair of MC6x18 channels, I can double the moment of inertia, and the max bending stresses will actually be a little lower than it is now (theoretical), but it's the deflection that's primarily of interest. Modelled as a continuous beam, the max deflection at 50 psf would be only .087 in, which in a span of 186 inches (one half of the 31' total), yields L/2123, while the original is about L/2880.
Modelled as a pair of simply supported beams, max deflection at 50 psf is .21 in, which gives L/886. The existing beam is continuous, though reinforcement would not have to be, and could be broken at the locations of zero bending stress.
The 'bounciness' is probably best modelled as a point load near mid-span (between column and wall) on a continuous beam, and the longer, reinforced span would be about 60% as stiff as the original, with it's shorter span. But the big question (for me) is how stiff is considered the minimum for a solid feeling floor.
Someone expressed uncertainty as to the geometry, so here's a better (I hope) description. It's a rectangular space, 31' long by 22' wide. A continuous W8x18 steel beam runs the the long way down the center (31 feet), and floor joists run perpendicular to the beam. There are two columns under the beam, making 3 equally spaced spans. There is a wall on the floor above, supporting ceiling joists on one side only - the other side is trusses that span the 31' and support the ceiling, with no bearing on the wall or beam.
I want to put a column under the mid-point of the beam (with new foundation, or course), and remove the existing two columns. Beam reinforcement will be done, based on accepted criteria for deflection (bending and shear strength very low for any configuration) and 'bounciness'. The only thing I don't know is what the accepted criteria are.Be seeing you...
T,
"how would you reinforce it if needed?"
Epoxy or bolt a shallow "C" into each side.
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SamT
TKanzler ---
"I'll seal the drawings myself."
You don't know the right math to do. I would reject your calculations but ...
I don't know what the loads are. You don't know what the loads are. It does not really matter as we are both engineers.
I do symbolic computatation rather than numeric computation, but I will try to be helpful.
1) We will assume that the current beam is sufficient.
Look up the deflection formula for a beam simply supported at one end and fixed at the other. Divide by "L". That gives you the delection ratio.
The formula is correct for the current situation.
You new situation has "L" in that formula replaced by "1.5L"
Take the ratio of the 2 formulas. I expect the result is 1.5^3 (about 3.4).
I don't think you can add that much material.
What?????????????????????
You must not be an engineer.
Hello GHR.....we meet again.
Graduated high school at 14 after completing two years worth of pure and applied "A level" math (that's right....Calculus....University level Calculus). That was 3 years before the degree in Architecture.........etc.
In over 24 years in this field, I have calculated, reviewed, defended, and inspected more structural engineering for residential architecture than I care to recall.
I have NEVER however, seen anything even remotely similar to, or in any way resembling, the rambling "non-speak" of your post.
If you sir, are a practicing structural engineer, I strongly recommend that you either re-word your posts so that they demonstrate you expertise more efficiently, or kindly refrain from offering "engineering advise" in a format that is incomprehensible at best.
Good day.
I think GHR just likes listening to himself "talk". So far he's never contributed anything useful to a discussion on this board that I've seen. I was kinda impressed with that post, though. At least he finally showed he can do math, even if it didn't make any sense. (-:
I wouldn't have turned out the way I was if I didn't have all those old-fashioned values to rebel against. [Madonna]
Yeah....I know.
He always seems to find both of us at the same time. I wish that once, just once, I could learn something useful from his posts.
ps (other than .....he is an engineer........ which he tells me all the time).
"I wish that once, just once, I could learn something useful from his posts."
Same here. That's why most of us are here - To learn and to teach.
He seems to be the only one I can remember around here who just wants to pat himself on the back.
Our family once had a dog that was a mix of Pit Bull and collie. After it tore your arm off, it would go for help.
This is why I stop short on doling out structural advice. Even the best description by another structural engineer will not tell me everything I need to know to comfortably address the situation. Every situation is unique and what worked on the last ten jobs may not be feasible, or even cost effective, for this situation....that's not a mistake, it's rustic
"You don't know the right math to do. I would reject your calculations but ..."
Come again?
"I don't know what the loads are. You don't know what the loads are. It does not really matter as we are both engineers."
I was using a beam fully restrained at one end, and simply supported at the other, same as you mentioned above, with a distributed load based on 40 + 10 lb/sf (and the dimensions of the floor area given before). Bending stresses are not much better than simply supported, but deflection is certainly better when analyzed as a continous beam. I don't know what the required live and dead load criteria are, and I will talk with the town to ascertain that (they are very easy to work with, and not like the knuckle heads found in many building departments).
Deflection from a walking person is probably more accurately modelled as a point load on a continuous beam with three supports (right out of the AISC ASD manual, or Roark's, or any number of other references), but that doesn't take into account the dynamics (mass and stiffness), though I doubt that too many people take it that far, especially on a house like mine.
"1) We will assume that the current beam is sufficient."
I'm sure it is. But I don't want to just scale up what's there, since I don't know what the criteria were when it was sized, and therefore how close to the minimum it is. I started by looking at the same deflection, and got a very large beam as a result (1.5^4), and that started me wondering just what is actually required.
But many, many times I've seen people talk about giving the design to their supplier, and letting them size a beam for them. That can make folks comfortable, since someone else says it's OK, but I want to know what the requirements actually are, see how that compares to what I've got, and from there go forward. If this stuff is normally pulled from a cookbook, then I'd be curious to know what the beam properties would be for the given conditions with just one column.
There was a discussion here many moons ago about floor dynamics, and someone posted a link to a good dissertation on the subject (Boss Hog maybe?). Anyone have any recollection about that? I should have copied or bookmarked it, but I didn't anticipate doing this work at the time. Be seeing you...
Usually we use deflecion/length. That explains the difference between my 1.5^3 and your 1.5^4.While floor deflection ratios (deflection/length) are held to 1/480 or 1/360, girder deflection ratios are held to much less. Your numbers are not unreasonable.If you keep the deflection ratio of the new girder the same as the old, the vibration of the new floor will be the same as the old.Reddy's book on composite plate theory provides an accurate formula for computing vibration of floors. I point this out because the locals on the board use a formula that is in error.