Dimensional lumber load carrying ability

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Does anyone know if there is a place on the web that shows the load carrying ability of different sizes of dimensional lumber over different spans?
It sure be helpful if someone does.
Thanks
j walker
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Try http://www.fpl.fs.fed.us/documnts/fplgtr/fplgtr113/fplgtr113.htm
Everything you asked about - and more...
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J,

This is a good one. Be sure to select "US Species" if you are in the US.
http://www.cwc.ca/design/tools/calcs/SpanCalc_2002 /
Anthony
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Thanks for the links. Now I am off to find an engineer to interpret the tables for me.
Thanks again.

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Hi J Walker, <--gotta love that...
Try these online calculators:
http://www.woodbin.com/calcs/index.htm
The one you may want is called the sagulator, about the 6th link down.
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Mike Wenzloff
Wenzloff & Sons Cabinet Makers
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Thanks for this reply. This site is easy to use for shelves but I found the terminology when applying the application to a beam to be difficult.
Thanks again!
On Thu, 14 Apr 2005 00:58:45 GMT, "Mike Wenzloff" <mwenz *@* wenzloff.com> wrote:>Hi J Walker, <--gotta love that...

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j walker wrote:

Just make the "depth" equal to the width of the beam and the "thickness" equal to the height.
A shelf is just a "deep", "thin" beam. A beam is just a "thick" shelf that is not very "deep".

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I think that I use the term "beam" in a different manner.
I call a floor joist a beam and apparently that is incorrect.
This clarifies one thing; why the sagulator refers to a beam "standing on end."
So, how do I get the sag in a floor joist?
What I am attempting to understand is how much sag should occur in a 2X8 floor joist if the span is 8 feet. The application is for a water tank stand which will measure 8X8 and will hold an 1100 gallon water tank. The rim joists are 2X12 and the floor joist ends rest on a ledger nailed to the side on the 2X12. There are 9 joists in between the 2X12.
On Sat, 16 Apr 2005 22:50:20 -0400, "J. Clarke"

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j walker wrote:

A floor joist is a beam.

To the sagulator a beam is like a shelf stood on its edge in a manner of speaking. Remember, the sagulator was intended for shelf deflection. The math is the same for anything rectangular in cross section and of uniform material properties, what is different is the words one might use to describe different measurements. "On end" is I suspect careless terminology--I suspect that "on edge" was meant.
To the sagulator, looking at your beam from the end, (i.e. so that you're looking into the end grain in my pathetic attempt at ASCII art below)
(this is the top, the load you are supporting would be up here) ____________________ __ | | /|\ | | | | | | | | | | | | | | | | | | | | | | | thickness | | | | | | | | | | | | | | | | | | | | | |____________________|_\|/ |<----- depth ------>|
And the "width" in the sagulator is the "length" of your beam (i.e. 8 feet, only expressed in inches).

If I read you correctly there are 9 2x8s supporting this tank, with the 2x8s being supported on the ends by 2x12s.
To use the sagulator to get an approximation of the sag in the 2x8s, you would plug 1100 pounds as the weight, uniform load (I'm assuming that the tank covers the full span--if you want to be safe call it a "center load", which will calculate a higher deflection), 96 inches as the width (8 ft x 12 inches), 18 as the depth (2 inches per beam x 9 beams), and 8 as the thickness. You'd also of course have to pick the appropriate species of wood. You'd then want to do the same calculation for the 2x12s to make sure that they're up to the task.

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J. Clarke wrote:

My apologies I misread "gallons" as "pounds"--that should be 8800 pounds (1 gallon weighs 8 pounds to a good approximation--8.33 to be exact).

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"J. Clarke" wrote: ...

But the calculation is on each joist so the load/joist is only 1/9 (or 1/8 if one wants to discount an end and be a little more conservative) so the weight to plug into the robot is roughly 1000 lb (or 1150 lb if use 8 instead of 9).
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Duane Bozarth wrote:

Note that rather than dividing the load I multiplied the width to encompass all members, so for the method I decribed 8800 would be the load to apply.
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"J. Clarke" wrote: ...

Sorry, I missed that...
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j walker wrote:

What you mean? Different from what?

Why you think that? It functions that way, a "joist" is a specific use.

A beam "standing on end" is a column.

I plugged in 1100 gal * 62.3 lb/ft^3 * 0.1337 lb/gal / 9 joists ==> 1020 lb load per joist. Putting in a span of 96", height of 7.5 and depth of 1.5 for an tuba8 and a long-leaf pine (SYP) for specie I got a deflection of roughly 0.1". For a 2x10 it was closer to 0.05.
Depending on your actual orientation/layout, it might be better to consider the load distributed on 8 instead of 9 intermediate joists. That'd raise the load by 9/8 and increase the deflection to 0.15", roughly for the 2x8.
HTH...
BTW, seems reasonable altho I didn't hand check the calcs returned by the robot...
What you need to make sure of (at a bare minimum) is that you've got adequate structural connection details throughout the <entire> structure, including whatever the 2x12's of which you speak are resting upon and what their span/spacing is to ensure they have enough additional loading bearing capacity.
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Duane Bozarth wrote: ...

...
That of course, assumes uniformly distributed load...I just noticed after posting the "stand" part...how the stand is configured determines how the load is transferred to the floor and whether the floor can support the concentrated load if it has small contact points is yet another question...
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"Duane Bozarth" wrote in message

<snip of much good advice>
... and the 'crown' always goes up.
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Duane Bozarth wrote:

...
BTW, don't neglect the weight of the tank and other structural material as well as whatever other dead load there is plus there should be a safety margin as well.
And, for the supporting 2x12 and remaining portion of the structure, don't fail to include the other loads they're required to support as well--which may not be simply determined w/o someone much more expert looking at the design...
Remember the weight of this tank/contents is <additional> load beyond other structural loads.
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I would recommend that you have an engineer do your load calculation and framing plan for this.
Keep in mind a couple of things:
The liquid volume alone will weigh about 9160 Lbs. before you add anything in for what will hold the water, temporary point loading during servicing, base level allowance for live load, etc.
Let's say that your tank and water come in at around 10,000 Lbs. (we'll leave out the other stuff for now).
You are proposing to carry this on eight joists which will rest on a ledger. Assuming the ledger to be 2X stock (1-1/2" net) you will have 36" to rest your 10,000 Lb. load on (1-1/2 x 1-1/2 x 16 = 36).
That is about 278 Lbs. per Sq. In.
Let's think about the fact that you are using a 2 x 12 rim joist and butting 2 x8 joists to it on a ledger.
The 2 x 12 is 11-1/4" and the 2 x 8 is 7-1/4", leaving 4" for the ledger. Unless you specify that the ledger must be ripped from a 2 x 6 to the full 4" height, you may wind up with a 2 x 4 ledger sitting 1/2" above the top plate. I wouldn't want to see the fasteners handle that load all by their lonesome. Neither would I want to see the space shimmed with typical shim stock.
I would want to make sure that the plate was doubled and that the joists rested directly over the vertical framing members below the plate, and that the load was transmitted all the way to the foundation in the same fashion.
It seems to me that the amount of sag is not the real problem. That is not to say that it is not a problem at all. I would glue and screw 1/2" min. ply (not CDX) to the underside of the joists and glue and screw the subfloor on the top, creating a sort of torsion box on the cheap.
The other thing that I would consider is that you are attempting to hold a lot of water on a framing system that will decline in its ability to hold the weight if it is subjected to water damage.
Anything that holds water will eventually leak.
These sorts of considerations are what make people like me pick up the phone and transfer the liability to my friendly local engineering firm.
Tom Watson - WoodDorker tjwatson1ATcomcastDOTnet (email) http://home.comcast.net/~tjwatson1/ (website)
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wrote:

Tom,
Why isn't the calculation above (1-1/2 + 1-1/2) x 8 = 24

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Because you have eight joists with two ends apiece.
This gives you your theoretical point load.
The most important thing that I said in the post, and why I repeated it, is that you should get an engineer in on this.
Clarke and Bozarth have responded with math that does not take into consideration the stresses and loads that are involved in this.
What you are describing is an unusual load situation that is beyond the capabilities of most onsite guys. I would want to be very particular in my inquiry that the stresses could be absorbed all the way down to the foundation level.
You are contemplating sitting a huge mass of potential energy in a position that can do great damage if things go wrong.
Good luck with it.
Tom Watson - WoodDorker tjwatson1ATcomcastDOTnet (email) http://home.comcast.net/~tjwatson1/ (website)
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