shed

thedeeredude · September 18, 2005, 03:04:45 PM

My dad and me are going to be building a post and beam shed. We've settled on using concrete piers in ground to hold up the frame, but one question. I've tried using the calculators in the forum toolbox to figure out how much weight can be held on the beams but have no idea how to work those calcs. So the question, would 4x6 white oak beams spaced 4' apart with a 7' span and 1.5" white oak decking be enough to hold a small tractor(farmall cub sized)? And can you show your calculations so I can figure out how to do this myself? Thanks a bunch.

Don P · September 18, 2005, 10:58:05 PM

I've got some other beam calcs here, your loading is a little different than say a house floor;
http://www.ls.net/~windyhill/Calcs/beamindex.html

I know nothing about the beams supporting the joists, their size or span, this is just the 4x6 joists themselves. No other loads have been figured (like the joists and decking itself).
I would guess your situation would be point loading at third points; I'm a carpenter not an engineer so you're getting cheap advice, treat it accordingly ;D ;).
http://www.ls.net/~windyhill/Calcs/2ptbeam.html

a rule of thumb for something like a tractor with its rolling loading and braking (brake gently!) is to double the load. I'm guessing and giving every tire 1000 adjusted pounds figuring the machine might go a ton then doubling for dynamic loading and dividing by 4 wheels...adjust accordingly. Either way I put the tractor on the joists, in my mind I come up with pretty much the same loading condition. So 1000 in the load box.
Span 84
4 wide
6 deep
..from the design values popup..
Max Fb,1200, #1 White Oak
MOE 1
Fv 105
I looked at deflection again under what I assume is the resting load, 500 pounds. Deflection gets back under the limit, it does bounce under the dynamic "fudge" load pretty strongly and approaches the allowable bending strength limits.

Let me know where I'm confusing you if that doesn't get it.

I need to read up on the allowable design values and flat use adjustment values to figure the decking,
That said, I ran it real quick on this calc for single center point load ,
http://www.ls.net/~windyhill/Calcs/beamclc_ctrptld.htm
I would figure one 1000 pound load at midspan on one plank. See if you get 2.75" x 8"

Those calcs are really just doing the number crunching that appears in engineering texts and the Nat'l Design Specification for Wood Construction. I put the formulas in those Javascript calculators so I wouldn't have to find the formulas or make those dumb little math errors halfway thru a long string of cyphering that I'm so good at :D. I hope they can be of some use. If I've totally muddied it up, let me know and I'll post the formulas themselves...I might be making it harder than it is

.

Don P · September 20, 2005, 09:56:07 PM

I must be making it harder, here's the longhand.

You need to check a beam's size and grade 3 ways. In bending, in deflection, and in horizontal shear.

Using the example of a beam loaded at third points the math is like this

Bending;
M (Maximum bending moment)= P (point load) x L (span length (feet)) / 3
M= (1000 *84) /3
M=28000 in-lbs

Now you need to figure out how large a section modulus you need for the bending strength of wood you have;
Section = M / Fb
S= 28000/ 1200 psi (Fb of #1 WO)
S= 23.33"

Now you check your beam to see if it has the required section. Notice in this calculation that depth is heavily favored as the way to pick up bending strength fastest.

S=( b * (d * d)) / 6
S= (4*(6*6))/6
S= (4*36)/6
S= 24

Congratulations, your proposed beam of 24" is slightly larger than the required 23.3" one, you pass in bending.

Next check Deflection;
D=23P(L cubed)/648EI
E is Modulus of Elasticity, I is moment of inertia (width of beam x depth cubed)/12

D= 23*1000*(84*84*84)/648*1,000,000*72
D= .292" of deflection
Code says deflection in a floor must be less than 1/360 of its span (a 360 inch beam can sag 1 inch)
84/360=.233 allowable deflection.
You sagged .292" more than the allowed .233 it barely fails in deflection.

Third check, Shear;
V (shear)=P
V=1000
Fv(fiberstress in shear) =(( 3 / 2 )* V ) / ( b * d )
Fv= 1500/24
Fv= 62.5 psi
Allowable shear is 105, you pass.

Is this helping at all?

Jim_Rogers · September 21, 2005, 03:03:53 PM

Don P:
Nice review of the formulas. :P

Looks like you've got it figured out ok.

Don't you usually use grade two or better values?

And when a beam is more than 2" larger in depth than width you would normally use the Beams and Stringer values, don't you?

Jim Rogers

Don P · September 21, 2005, 10:54:33 PM

Thanks Jim,
I was hoping you would look it over. I did push the grade up to #1 Beams and Stringers to make the joists work and should have noted it loudly

This is nice stock, the #1 the Math needed is Fb 1200 psi, a #2 that is normally called is Fb 750 psi and an E of .8. In #2 B+S it would need to be a 4x7.5" deep beam. I guess technically at 4x6 we should be using the dimension lumber values (2"-4" thick), Fb 1200 is Select Structural W.Oak there, close to clear lumber.

This isn't the way I would build this floor, but it is how to do the math to model how things work. Running through it real quick with the calculator doing the math lets you concentrate on the changes that bring a beam into range. At 4' joist spacing and high quality joists everything was barely acceptable. Then I worked the decking and it comes in too thick. That tells me it shouldn't be spanning so far. It also tells me that by the time I put that much deck weight on the joists, they probably won't pass.

If the 4x6 joists are spaced on 2' centers then the decking comes in at 2x8 and all materials work at #2 grade. Because you then fall into dimensional repetative use adjustment (15%) and size factor adjustments (30%), the 850 psi allowable bending design value of "dimensional" #2 W.O. becomes 1270 psi. The joists got closer together and are at the limit of where they are considered to be sharing some load from each other.

Flat use factors for laying a 2x board flatways as in decking, multiply the bending value, Fb, times "flat use factors". A 2x 6 or 8 is given a 15% increase over its base design value, a 2x4, 10%. A 2x10 or wider, 20%.

thedeeredude · September 22, 2005, 09:35:45 PM

Well,
Thanks for the replies fellows. We might have to revamp the design due to township codes and the like. What I'm thinking is just using 3x6"s 2.5'OC with 2x6 decking. I have absolutely no idea how you figure that stuff out DonP, but thanks for helping me out.