You would need the advice of civil engineer for this.
To just ballpark it google some info on weight per cubic foot of concrete, load per foot of span for the beam, etc.
Do a little math and bingo, you will have an approximate idea of what is required.
Don't forget to include the weight of anything on top of your structure. Like dirt, snow, vehicles or other structures.
Good luck with it!

 

I think a structural engineer is probably who you should talk to, not a civil engineer.

 

Is it temporary until the concrete is cured, or is the beam supposed to be permanent?

If it's temporary, you could just use 2*6 lumber supports with a couple jacks or vertical kickers, but 4" seems thin for a cantilever.

 

12 ft by ???

 

I work with a bunch of engineers, I'll try to get you an answer, do you have plan for the beam spacing?

 

I'm no engineer but I suspect your corrugated panels would collapse long before your I-beams did regardless of spacing. Engineers feel free to correct me if I'm wrong on that.

 

Roofs are made of them capable of bearing 100lb per square foot of snow load. I've walked on them, and accidentally not stepped on the rafter; I didn't fall through. That's in the ~5" thick range for concrete. Using thicker material made for use as concrete forms would hold much more.

 

OK, I'll pick your brain, I'm an EE so don't talk over my head . The number 6 in your beam designation is the height and the number after the X is lbs of steel per foot? What would you use for a 20' X 20' concrete block underground structure (all holes filled), with 3 I beams across at 5' spacing, with a 6" reinforced slab on top, no snow load, no dirt load as it will be the floor of a ramada, I was thinking a W12Xsomething but I really have no clue.

Rancher

 

I'm in the concrete construction business and I have built many storm shelters, covered basements and cellars over the past 35 years, most steel yards will have an onsite engineer who will tell you what you need,
The smallest I beam I have ever put under anything was 1o inches tall by 4 inches wide by 3/8ths . You will need wooden beams[2x6 doubled] every 2 ft lengthways with upright legs every 2ft min. for support until concrete cures.
You should have a rebar mat of at least no2 bar on 12inch centers, a double mat is better. I would suggest adding Helix[stainless wire] I love this stuff an its tough. It can replace rebar in a structure according to the engineers.
What you are wanting to do is not a problem, you just want to do it right and safely. Good luck with your project, if I was near you I'd give you a hand.:thumb:

 

You should have a rebar mat of at least no2 bar on 12inch centers, a double mat is better. I would suggest adding Helix[stainless wire] I love this stuff an its tough. It can replace rebar in a structure according to the engineers.

Please explain Helix, I was planning on a #4 bar 12X12 re-enforcing, you never see #2 around here.

Rancher

 

A quick rule of thumb to calculate the necessary depth of a beam, divide the span (in inches) by 20. For example, a 20' span would be: (20x12)/20 = 12"

The width of this beam would be between 1/3 and 1/2 the depth (4" - 6").

The dimensions of a girder would be the same, but the flange would be thicker. A beam overhang can be a maximum of 3/8 of the supported span.

 

A quick rule of thumb to calculate the necessary depth of a beam, divide the span (in inches) by 20. For example, a 20' span would be: (20x12)/20 = 12"

The width of this beam would be between 1/3 and 1/2 the depth (4" - 6").

Got all that, what about spacing of the beams? Formula for that? Or at least a recommendation?

The dimensions of a girder would be the same, but the flange would be thicker. A beam overhang can be a maximum of 3/8 of the supported span.

Not understanding this part.

 

The plans are to build a 24x32 ft basement that is also divided down the middle, so there will be three 32 ft walls and two 24ft walls at each end. All of the walls will be filled and re bar reinforced. At the placement point of the beams, the walls will be extra reinforced. I was thinking that I could place the beams the 24ft distance on top of the walls at about a 2ft spacing, then put down the heavy gauge tin on top of that. Cover with 4 to 6 inches of concrete. This will be the roof for the basement and the floor for the shop above it. There will be no vehicle traffic, just a few freezers, air compressor, and woodworking tools.

 

... I was thinking that I could place the beams the 24ft distance on top of the walls at about a 2ft spacing, then put down the heavy gauge tin on top of that. Cover with 4 to 6 inches of concrete. ...There will be no vehicle traffic, just a few freezers, air compressor, and woodworking tools.

There is no reason to have those beams closer than 6' if you make the concrete thick enough to support itself. You're only talking about a 12' span, that should be easy to achieve. If you went for a thickness appropriate to a semi-commercial structure, around 8", you would need far fewer beams (I'm talking for the final structure, not the pour).

Some quick price checks put a beam of that size, weight and length around $270 ( 24' W6x15 undelivered). You could make a 24'x12' area 2" thicker (that's a little less than 2 cu yd) for less than the price of one beam.

I'm not sure what type of construction you were planning for the shop, but if it is going to be stick frame, you could build the supports for the pour yourself from lumber, and reuse it after the concrete is cured. Otherwise, just rent/lease some jacks for the pour, and temporarily back your sheet metal with plywood.

 

If you do not temporarily support that corrugated tin you will have a basement full of liquid cement. That stuff is tremendously heavy and has absolutely no strength till it sets up. If your beam is supported the entire length by walls you can get by with even an 8" tall beam. If not then adjust accordingly. You get more strength from height than width. Would place two runs of 4X6s supported by 4X4 posts the length of each room to support the wet cement's weight. That means you are supporting it every 4' the total length of the basement. When completely cured(would give it a couple weeks) you can remove the temp supports. Ask anyone who has ever poured concrete and they will tell you of "busted" forms and lost concrete. Over kill is the rule of thumb when shoring the stuff. Don't forget utility connection holes/pipes but weld them in place. When pouring do not let too much get in one place before spreading. Have plenty of help and plan on staying with it for several hours to finish the surface with what ever design desired. Set your building bolts well before it starts setting up. Saves boring later. Preparation and planning will save you lots of trouble/time later. "Skree" boards and "bull floats" are your friend with this much. Do not "scrimp" on your shoring(verticals)--remember--that stuff is very heavy.

 

06 has it right, I'd rather put in too much bracing than find out I didn't use enough. One hit to your contractors liability insurance will make sure you don't screw up twice, an you dang well wont hire no dope smokers ever again. That was my own fault, trusting that the job would be done right without visually checking myself.

Helix is stainless wire about 2inches long, it takes about 4 lbs to the yard , it is similar to fibermesh but instead of plastic it's stainless. You just dump it in the mixer and turn the drum about a min of 30 revolutions per yard if I remember right, If you want to know more about it I can put you in touch with a gentleman who has the franchise for it here in MO, He could answer your questions far better than I .

 

#2 bar is not really used and generally refered to as pencil bar. #3 =3/8" bar, #4 = 1/2" bar an so on. # 8 is 1".

 

You are correct about the no2 bar I should have said no4 bar. Reckon my thoughts outran my typing skills.

 

Crap! That happens to you too?:thumb:

 

I've been called old on this forum.

But the smallest they sell here is #4, and the largest is #6.

Just bought 10 #6 at $13.+ each to anchor RR ties into the ground.

Rancher

 

I've been called old on this forum.


Rancher

*frown*

Not nice, Mr. Rancher.

 

As a general rule, floor/roof decking materials don't hold didly without a network of closely spaced joists and beams to support them. To say a piece of plate steel will handle 2000lbs per square foot makes no sense. 2000 lbs/ft^2 may be the design load for a particular building which will dictate the design of the supporting structure. If you are building an underground shelter, your design load will take into account the weight of the concrete, dirt, snow, rain, ice, people, vehicles, etc. overhead. Don't forget the weight of the structure, itself.

Beam calculations can also apply to decking materials when you plug in the appropriate moment and modulus values because decking material is basically a really crappy beam. If you double the spacing between beam supports you will get 8 times as much (L^3) deflection for the same weight. If you double the width of a beam it only holds twice as much weight but if you double the height of a beam it holds 8 times as much weight for the same deflection or deflects 1/8 as much for the same weight. Since decking materials have very little vertical height, they are very flimsy. Load bearing doesn't go down quite as fast as deflection would indicate because the amount of deflection tolerated with load is proportional to length so load bearing goes down with the square of length rather than the cube. But that is total load. Distributed weight per square foot increases with length so you may be back to length cubed. Allowed deflection: Rafters length/180, floors length/360 (but often spec'ed at length/480 because people want floors). It also matters if the beam is allowed to flex or is rigidly supported at the support points. In practice, a 1" decking plate probably supports 16 times as much as a 1/4" decking plate. As you spread the supports, the thickness of steel plate required will rapidly increase to 1" and beyond. But you can get much more support for the same weight of steel by putting a bunch of beams under it rather than using thick plate.

For purposes of calculation, you should assume your beams/decking are not ridgedly supported unless you can prove otherwise. If a beam sits on top of a support column, it is not rigidly supported. If the bottom is attached to the support column but not the top, it is not rigidly supported. If the top and bottom are welded to the support column, then it is rigidly supported but there is a catch. The rigid support reduces the bending of the beam but it transfers those bending stresses to the column causing the column to bow and thus buckle easier. If the beam or decking is welded both sides of a rigid support column, it is also rigidly supported.

Strength mostly comes from having a top and bottom skin of the material with an internal structure that keeps them from shearing relative to one another. Imagine taking a ream of paper. support it on the edges with some bricks. Now press down on the center. It bends easily because the sheets of paper are free to shear. Now glue all the sheets together. Now you effectively have a piece of wood and it is much stronger. Hollow beams (rectangular tubing or wooden box beams) and I beams take advantage of the fact that you get more strength by taking the material and moving it away from the center, within reason, while still having a web of material to lock the top and bottom together.

You can model a simple floor as two layers. One layer is the joists and the other is the decking. Do the beam calculations for both. If your joists are themselves supported on beams between upright supports, then you have another layer to calculate. You also have to consider the compressive strength and buckling of your uprights or walls. You wouldn't want cinderblock walls crumbling under the weight of your joists and the weight they are carrying, for example.

You also have to consider concentrated weights. A car tire. The leg of a person, the leg of a piece of equipment, or a vehicle tire which puts the weight of a heavy object in one or a few places. This has to be added to your worst case design weights and you have to consider if you can punch through thin decking.

The taller the support columns (or walls) are, the more prone they are to buckling. Double the height and it will only take 1/4 as much weight to buckle them.

Trick question: how much does a 200lb person weight? If they are walking or jumping on your floor, it is way more than 200lb. How much more depends on the rigidity - how fast is their momentum decellerated?
In the event of an earthquake effective weights, such as the weight of dirt on the roof, could also be greatly increased because it is in motion. In an earthquake, your structure may not be evenly supported from below. As it rides the wave, it could be effectively supported only from the ends one instant and only supported in the center the next. The mass of your structure is also trying to stay still while the ground underneight it is moving side to side. If the structure is buried, you also have the mass of the earth beside the structure pushing it sideways. Most of the assumptions on which your calculations and design were based are invalidated.

If your structure is above ground, there are wind load issues. On a windy day, in a microburst, in a hurricane, or in a tornado there are substantial wind loads on your walls and other parts of your structure trying to topple it or buckle your walls. A tsunami or flood can be even worse. What I wrote about loads on roofs/floors also applies to walls. Walls are generally supported by the walls they join with perpendularly plus some floor and ceiling. Double the length of a wall and it may only take 1/8 the wind load for the same construction if you only consider support of adjacent walls. And wind load itself increases with the square of wind speed. This is one reason you see hurricane/tornado shelters having very small dimensions, such as an 8 foot cube. The walls and ceilings are supported by adjacent walls/ceilings such that the maximum distance to a support is 4 feet. Even so, each wall can see a force of 5 tons. In a 200mph wind, each 4x8 sheet of plywood on your exterior walls sees a force of about 5120lbs. A 32 foot long wall 8' high sees 40960lb and a double story structure twice that. Not counting roof.

Engineers and architects also multiply design loads by safety factors.

Crappy welds or inadequate fasteners can also cause collapse. Even good welds/fasteners if the design puts too much stress on them.

Here are some sites that have much of the formulas and data you need to do some basic calculations:
http://www.forestryforum.com/members/donp/beamsizing.htm
http://www.faztek.net/technical.html
http://www.engineersedge.com/calculators/section_square_case_4.htm
http://www.engineersedge.com/beam-deflection-menu.htm
http://en.wikipedia.org/wiki/Buckling
http://en.wikipedia.org/wiki/Deflection_(engineering)
http://bulk.resource.org/codes.gov/

If you do your own design and calculations, have it reviewed by someone qualified. Structural collapse is a deadly issue and a serious possibility, even in less than disaster situations. The information here can help you evaluate a concept or cost out a project. But there is a lot that has to be considered in an actual design before it is built.

Consider for a second an intermodal cargo container. Properly stacked, these can support hundreds of thousands of pounds. But only if the weight is where they were designed to take the weight. Piling dirt on top, does not cut it. Something many people don't realize when they think about burying them.

An excellent post from this thread:

http://www.survivalistboards.com/showthread.php?t=237290

 

this will give you an idea on what is available

www.saginawpipe.com/steel_i_beams.htm


Personally I would look into a thicker roof and include the beam into the concrete itself. Holes can be cut or drilled into the beam itself so you can push the rebar through it perpendicular.

 

The spacing of the beams is dependant on the guage of metal pan that you use. If you were to space beams at 8' O.C., you would have to size the pan accordingly. I would highly advise against using corrugated roofing as your pan. It is not designed for the application you are describing. My recommendation is to look up the International Residential Code on-line. It is perscriptive for structures, meaning if you follow your options on a chart it will advise what structural members to use. A has been said before over-shore the work until the concrete reaches it's prescribed strength, that is 28 days. If you are not cantilevering the concrete, I would recommend reinforcing with #3 rebar at 18" O.C each way. Almost forgot, you will need to determine your tributary area and the weight associated with it. Lets say as an example you were going to space the structure at 8'O.C. and span 12', The area of construction that would bear on the beam is 4' on each side of the beam times the length, o, in this example 8' X 12' = 96 S.F. of trib area. If the weight of your construction is 20lbs P.S.F. your load would be 1920lbs, that is the beam you are designing for. That is simple loading. If you have any point loads like a post above the beam it really complicates things. Another option is a post-tensioned hollow core slab. They would be custom made to your project and require nothing but solid perimeter walls to support them. The only drawback is they require a crane to place. T-Pac manufactures them here in Tucson, I am sure there are others throughout the country.

 

The spacing of the beams is dependant on the guage of metal pan that you use. If you were to space beams at 8' O.C., you would have to size the pan accordingly. I would highly advise against using corrugated roofing as your pan.

Are you sizing the pan on no additional support during curing of the concrete or...

I would recommend reinforcing with #3 rebar at 18" O.C each way.

You state you are in Tucson, I can't get #3 rebar just 80 miles South East of you...

Rancher