Structural Products (Lumber)

You might find info here:TDL - Structural Products - Lumber

American Wood Council (AWC) has "Span tables and Calculators" free online and may answer other ? you have

but the units of measure used was not something I could totally apply......

Will a good free units converter help you:-

http://joshmadison.com/software/convert-for-windows/

Hi Sid,

What type of lumber you want to use? Soft or hard wood? Let say hard wood is your choice; It's maple, oak, beech or ash for example.

Let us know your choice and we can eventually participate, Gil.

Hi Sid,

Let's see if I can help you out, but I'll have to make some educated guesses and assumptions first.

Assume you can buy 2x6x10' Spruce-Pine-Fir (South), commercial grade at your lumber store....it's pretty much readibly available. So is Southern Pine, but it's only slightly stronger by a little amount. So for argument sake, let's use the SPF lumber as an conservative approach.

Allowable Stress Values per NDS: Fb=950 psi (single use), Fv=75 psi, Fcp=335psi and E=1.2E6psi

A=8.25 In^2, Sx=7.563 In^3, and Ix=20.797 In^4

and "d" distance from face of support = 5.5 In. ; L=10'-0" = 120"

Limit LL deflection to L/480 = 0.25 In.

Assume your shelf width is W=24 In. out-to-out. Also assume that you're using a 2x6 as well on the backside of the shelf. Assume 1/2" plywood shelf

No Live load Reductions or corrections....ie, long term loading

We must work backwards from the Allowable stresses to deermine the overall Uniform Distributed Live Load that can be placed along the entire span of the shelf (notice I didn't say a single concentrated load at the shelf mid-span? The reason why is to design the shelf as if you're going to load it up during is lifetimewhich is a more realistic approach.).

use a Dead Load of 4 PSF to account for wood weight.

Using Allowable Bending Stress of 950 psi; Since fb=Fb=M/Sx and M=wL^2/8:

Resisting Moment, Mr = Fb * Sx = 950 psi * 7.563 In^3 = 7,183.9 in-#

Allowable w (TL) = 8 * Mr / L^2 = (8 * 7,183.9) / 120^2 = 3.99 PLI = 47.9 PLF or 23.95 PSF

Allowable W (LL) = 23.95 PSF - 4 PSF= roughly 20 PSF

&

Using Allowable Shear Stress Fv of 75 PSI; Since Fv = (3/2 * Vd) / A and "d" = 5.5 In.

also, L/2 - "d" = 120/2 - 5.5 = 54.5 In. @ critical shear location...

Allowable Vd = (Fv * A)/1.5 = (75 psi * 8.25 In^2)/1.5 = 412.5 #

Allowable W(TL) = Vd / 54.5 In. = 412.5 # / 54.5 In. = 7.57 PLI = 90.83 PLF or 45.41 PSF

Allowable W (LL) = 45.41 PSF - 4 PSF = 41.41 PSF

&

Using Allowable Bearing Stress Fcp of 335 PSI; Since Fcp = R / A and "b" = 1.5 In. and assuming a minimum bearing length at supports = 1.5"

and since R= W(TL) / (L / 2)......

Allowable R (TL) = (335 PSI * (1.5 * 1.5)) = 753.75 #

Allowable W (TL) = R * (L/2) = 753.75 * (120/2) = 12.56 PLI = 150.75 PLF or 75.38 PSF

Allowable W (LL) = 75.38 - 4 = 71.38 PSF

&

Since LL Deflection = (5*W(TL)*L^4)/(384*E*Ix), work out Allowable W(LL) if Live load Deflection Limit = L/480 = 0.25 In.

Allowable W(TL) = (0.25 * 384 * 1.2E6 * 21.797)/(5 * 120^4) = 2.44 PLI = 29.31 PLF or 14.65 PSF

Allowable W(LL) = 14.65 PSF - 4 PSF = 10 PSF <<<<GOVERNS>>>

As you an see, with only using 2x6 SPF lumber, you'll be limited to only loading the shelf with 10 pound per square foot Live load using a shelf span of 10 feet x 2 feet width, based on the Live Load Deflection Limit. You can exceed the deflection limit and can load the shelf to 20 psf Live load, but the deflection will become twice as large, or 0.50 Inch.

You may want to consider supporting the shelving every 5'-0, then you can load it much more.......just follow my caculations and use the new span data of 60 Inches instead.

Someone, please check my math as I haven't done so due to time constraints on this end....but you get the idea if I screwed up *LOL*

Hope this helps you! have a sunny day!!!

Hi Capt,

Interesting but complicated answer for a so simple question. Any way, could you make the demonstration: When the SPF lumber is sitting on the 2" face, what will be the result and what calculation or formulation to use?

Appreciate your explanations, Gil.

WOW - far more than my puny mind can handle (let me take a drink of wine first!!)

The things I found on line were giving stresses in PSI (the unit I was not so sure with) I understand all you are saying, barely. But I guess I can't quite wrap my brain around load per square foot (I'm and electronics guy, I do bettter with volts and ohms).

Maybe for a little added detail, I was going to bolt the rear 2x6 to the garage wall, then come out 4 feet with a total span of ~18feet across the garage with a post in the middle. Guess I was looking to see if I could place my 200lbs on it with some boxes of Christmas lights, power tools, baby crib - unless I have a grandbaby soon - and the like. I could even use a ladder and not walk on it.

Little better, or clear as mud?!?!? Thanks for the brain power!! ss

Gil,

It's not as simple as you are making it out to be, and yet then again, in some ways it is simple. Let me please explain, okay?

In order to check to see if a piece of lumber that is simply supported at each end and uniformly supported the entire span, then you need to check 3 different stresses using the Applied Design Load and compare to the maximum Allowable (and corresponding) Stress. You must also check the Live Load Deflection @ midspan caused by the Applied Live Load against the Allowable Live Load Deflection Limit. All I did is work the problem backwards with the Allowable Stresses as a starting point, and used basic beam formulas to solve for th corresponding Allowable Design Loads, both Total Load and Live Load. Dead load was already a given amount based on the wood weight and how the shelf was to be built.......TL = LL + DL. with 2 parallel 2 x 6 shelf beams 24 inches apart supporting a 1/2" thick piece of plywood and supported at the end with a vertical post of some type and size......

For instance, normally one would perform the calculations as follows:

For Shear Stress: Design Shear Stress (fv) must be less than Allowable Shear Stress (Fv) for the beam to be acceptable. The formula for determining Shear Stress in wood is: fv = 3/2 * Vd / A, where Vd is the vertical shear value taken at the "critical location", which is equal to the depth of the beam, d, from the face of the beam end support.

For Bending Stress: Design Bending Stress (fb) must be less than Allowable Bending Stress (Fb) for the beam to be acceptable. First one must determine the Bending Moment for a simply supported beam that is uniformly loaded. This formula can be found in the beam tables of any statics textbook. M = (w * L^2) / 8. And Fb = M / Sx. Sx is the section modulus for the beam.... Sx = (b * d^2) / 6.

For Bearing Stress: Design Bearing Stress (fcp) must be less than Allowable Bearing Stress (Fcp) for the beam to be acceptable. This stress situation occurs where the ends of the beam bear directly down onto a contact area. Rarely does this stress govern in design, but must be checked in case you have a very large load and small span as well as a small contact area. This stress occurs perpendicular to the grain. fcp = R / Ab, where R is the beam end reaction due to the Total Design Load acting over the total span, and Ab is the contact area, which in our case our 1.5" wide beam was bearing on a cross piece of 1.5" width or a vertical post 1.5" wide. R = total load W * L / 2.

W is the uniformly applied design load.... W = (2.0' wide shelf / 2 load carrying beams) * Design Total load in Pounds per square foot.

Live load Deflection, Delta = (5 * W * L^4) / (384 * E * Ix); where W is the Design Live Load in psf; L is the beam span in Inches; E is the Young's Modulus of Elasticity in psi; and, Ix is the beam Moment of Inertia in In^4...Ix = (b * d^3) / 12.

I hope this has made things clearer.

Sid---- you're welcome. If you want to determine the bending stress in a beam with a single concentrated live load at midspan, and the uniformly applied DL of the wood beam:

Max. M = (P * L / 4) + (W * L^2 / 8)....where P is your concentrated live load.....and W is the weight of the wood in Pounds per Lineal Inch (PLI).

notice that I have been using "like terms" throughout the comps: Pounds and Inches. Don't mess up things using inches in one calc and feet in another!

Hi Moosie,

I am not making complicated. Also, I use different types of wood only to colour, inside and out. My question was; If we make a shelf, the lumber sitting on the 6" side. I asked; What's happening when we let the lumber sit as a beam on the 2" side? This was simple and the only addition.

Thanks for the teaching but I think we need some other important definitions before calculations. The type of wood we use, the glue we put to hold some suggestions together like SPF, etc...

All your teaching is fine and I catch it! You have 12 paragraphs and this is the ration between foot and inch, isn't it? You made that special for me? Thanks, thanks again from Gil.

Gil, this has the be real short as I have a Dr. appointment to go to.

My calculations were for the lumber on edge, or otherwise laying on it's 1.5' wide side with the 5.5" in the vertical.....the lumber is strong that way by quite a lot.

SPF means Spruce-Pine-Fir, a species of pine tree.

bye

Hi Moosie,

Forget the calculations, I already said that I catch everything!

I would like to get back an answer from you about the ratio of your paragraphs and the inch/foot ratio. It's possible in between all these confusions and explanations.

Another suggestion; Imagine, you have or the person who initiated the question, two lumbers in vertical position 11/2" by 6'' spaced by 6", and covered for the shelf by 1"x6" lumber. This solution will support more or less weight? Let me know but don't give me the mathematical solution. It will be different for different type of lumber, isn't it?

I already explained what it means SPF in an early comment! You have to read us at place of telling that we don't know.

There is no ratio between my paragraphs and yours! It's clear for me and should be the same for you. All the best for your the doctor's visit. Watch yourself because he will give you some pills for one of your sickness and these pills will develop another problem(s) to you. Poor man with health problems! Now, everything is in plural! Oh la la, watch around, Gil.

Gil,

I'm not quite getting where you're coming from.....about the number or ratio of my paragraphs and the inches to foot thing....nothing intentional made by me I assure you.

Forgive me if I'm a little fuzzy this afternoon after returning from the Doctor's appointment.....and a 3-hour round trip drive on top of everything else. Upon my return I've downed a few bottles of Guinness Extra Stout to celebrate the Irish in me and all of us. Me thanks me VERY LOVELY blue-eyed Red Headed Irish wife (1st Generation USA) for greeting this weary soul at the door with an opened bottle of the brown stuff!!!! Ahhh you should wifffffff the aroma of the Corned Beef, cabbage and tators cooking on the stove.....ahhhh I tell it's heaven on Earth here tonight!!!!

Gil, the arrangement of the 3 - 2x6's you gave me makes things much more complicated to analyze and determine the governing allowable load. I understand where you're coming from, but the calculations that I provided before will only aide you a little bit. Sure, it'll hold more weight. It's a difficult pickle because it's what we structural guys call a built-up or composite section. First you have to determine the overall sectional properties: A, Sx (top), Sx (bottom), Ix and y-bar (locates the neutral axis from the extreme top and bottom fibers where you'll have the largest bending stresses). Then you can apply much of the stuff I showed you guys how to do. Once you've determined the Governing Allowable Live Load, then you have to check the linear Shear Stress occurring at each joint, be it glued and nailed or screwed or a combination thereof. You need to do this to make sure the entire built-up section doesn't fall apart on you! OR fly apart as is most of the cases!!! The maximum shear stress at these joints between the 2x6s will occurs at each beam support....where Vmax = Reaction.

And yes, to answer your question the load carrying capacity of whatever beam you use (as well as sizes) will differ from one wood species to the next, and even between wood grades of the same species, because all of the Allowable Design Stresses are different......no two woods are created equal!!! Hiccup!!!! LOL

Hello Moosie,

I accept the first paragraph. You are good with the last apragraph too. The in-between is personal and no one want to read. I get what I need, and wish you a good and envigorating night, Gil.

Hi Moosie,

I understand your disappointment by my bad explanation. This clearly my error! Sorry enough to be accepted.

I forget to put down that the two vertical lumbers must be at 6" distance from each other but centre of the 11/2" vertical lumbers. I hope this is clear and succinct enough to be catchable by anyone!? Other way the 6" horizontal shelving lumber can eventually fall in between the two vertical, do you see it? Fine!!

Now, after my corrective action or intervention - it's good to know another languages, so you can add some delicacy in the conversation between two well educated person - and revisit your calculations, and check before you write down the final results.

Come on, Moosie! You gave the ratio to find between foot and inch or vice versa. So, I discover that you gave by making twelve paragraph, and you can check it, it was your comments, not mine, and make a final answer if my finding were successful or need to be corrected!

Please, don't copy my writing because other people could discover some failure and/or weaknesses of my English. However, this time I will accept this operation of cribbing others.

I am confident that the doctor did not give you some pills which can deteriorate your brain function. You know, about the neurons. Anyway, I see that you survived the visit and don't have any excessive personal dammages. Watch the number of pints or quarts you put behind the tie! You don't wear a tie? It's Irish Day?! It can be dangereous! I am sure you are enough in control of yourself and could make another answers, Gil.