On 5/12/2013 6:31 AM, Alexander Lamaison wrote:
> dpb<
no...@non.net> writes:
>> On 5/11/2013 3:05 PM, woodchucker wrote:
...
>>
>>> I would like to chime in here.
>>> I don't think the structure will be strong enough. I think the boards
>>> for the shelves are not oriented correctly on the bottom. The top has a
>>> skirt that helps but all boards in my estimation should be set on edge
>>> to hold the heavy weight of the logs. That way they won't sag.
>
> You mean flip the shelves 90 degree to make them stand like joists
> rather than floor boards? That would make them strong and let more air
> in (both good). How should I join them to the supporting beams without
> losing the strength of having them run accross the top, because now
> they'll be a bit thick to drive a screw through the top of them, even if
> I reduce the vertical dimension a bit?
>
> Or did you mean something else?
Yes, he means run as joists. That's certainly one way altho it will cut
down the opening size significantly unless you raise overall height.
You could support them simplest by adding a ledger board on the ends to
rest them on and only w/ a little more effort notch them to hide it.
>>> Your corner and middle supports can be downsized to conventional stud
>>> lumber. They are vertical and the stresses are less than you think. It's
>>> the shelves that need the most.
>
> Nice. That more than halves their cost.
Could even go to two 1x joined in tee and for the vertical will still be
plenty stout-enough...
...
> Yes, sorry, should have mentioned the dimensions. The shelves are 25 x
> 100 finished size (1" x 4" ish).
>
...
>>
>> If assume oak at roughly 45 lb/cu-ft, that's 1200 lb on each of those
>> shelves which translates to an average loading of ~135 psf. That's
>> pretty healthy load; more than I'd have thought. Of course, it's not
>> likely that the actual firewood will be so accomodating as to fit so
>> well, but it is a point...
>
> Now this is getting fun :) Some engineering.
>
>> Now one can go look at deflection tables and so on and make some
>> choices on sizes.
>
> I've not come across deflection tables before. Do you have an online
> reference to the one you're using? The ones I'm finding online are all
> for large structural timbers for house building and don't cover the
> kinds of pieces we're talking about here.
Well, all you have to do is set the dimensions correctly and use
appropriate values for the various material and geometric properties...
I use the beam calculator quite a lot...but for stuff like you're doing
the "sagulator" is probably the easiest tool as it hides a lot of the
complexity by making all the assumptions about material properties and
assumes rectangular pieces...
<
http://www.woodbin.com/calcs/sagulator.htm>
>> Just looking at a minimum I'd put a solid piece of 3/4 ply or the like
>> in the middle between the two supports
>
> I'm not sure where you mean. Across the shelves, knitting the 5 pieces
> together like another support beam?
Well, if I read your drawing correctly, it's open the full width w/ the
exception of there being a vertical support at the front and rear in the
middle but no support in the middle internally. Since there would be a
full 5-ft span w/o those that's obviously not possible/reasonable to
consider removing it; it will prevent any use except of the two "bins"
so you're not losing anything by putting a divider between them for
vertical support across the full width. Ply would be the easiest way to
accomplish that; solid verticals would also work. You could either use
the full length shelving and fit the verticals or use solid one-piece
verticals and separate shelves--your choice, same result.
>> plus their definitely will need to be support across the width of the
>> openings under the shelves.
>
> I was avoiding this in order to allow air underneath. Perhaps I can
> compromise using a smaller piece that both provides support and leaves
> an air gap.
...
One way would be an on-edge "X" from corner to corner fastened to the
bottom of the shelves w/ adhesive and screws to "stiffen-up" the shelves
just as does an edging or a table apron.
If I take your design to the sagulator I get (using one of the pines
that would be typical lumberyard material here; pick a species that is
somewhat like what you would have for your material obviously)
Shelf Characteristics
Shelf Material
Shelf attachment Fixed X Floating
Shelf load 200 per foot
Load units lbs
Load distribution X Uniform load Center load
Build shelves with less waste
Shelf span 30 in
Depth (front to back)
Thickness
[Optional] Edging Strip
Material None
Width See note # 10
Thickness
New Apply WoodBin lab correction? yes x no
Sag total 0.27 unit
0.107 in per foot
Target sag: 0.02 in per foot
Just for comparison I took to the
<
http://www.engineeringtoolbox.com/beam-stress-deflection-d_1312.html>
beam calculation engine I use quite a lot and while it takes a little
more effort, I'm always comfortable in knowing what actually happens...
For it for the same assumptions
Imperial Units
16.666 q - Load (lb/in)
30 L - Length of Beam (in)
.333 I - Moment of Inertia (in4)
2x10^6 E - Modulus of Elasticity (psi)
0.5 y - Perpendicular distance from to neutral axis X (in)
Unit Load - q : 16.7 (lb/in)
Total Load : 500 (lb)
Length of Beam - L : 30 (in)
Moment of Inertia - I : 0.33 (in4)
Modulus of Elasticity - E : 2000000 (psi)
Perp. distance from neutral axis - y : 0.5 (in)
Support Force - R1 : 250 (lb)
Support Force - R2 : 250 (lb)
Maximum Stress - : 2815 (psi)
Maximum Deflection - : 0.26 (in)
I get a max deflection of 0.26 instead of 0.27 -- pretty doggone good
agreement.
Now, that's based on the previously estimated 1200 lb total load divided
out to the average uniform load on a 1x4 laying flat. You can see the
difference if you turn it around on edge or change various other
dimensions and or loadings, etc., etc., etc., ...
As I had presumed initially from just gut feelings, the sag would be
noticeable but a 1x would likely be able to hold the load w/o actually
breaking but it's well under-sized that way.
BTW, the I for a rectangular section is bh^3/12 where b=base and
h=height. In English units it generally has units therefore of in^4.
For your 1x4 flat that gives 4x1^3/12 = 1/3 in^4. You can see why on
edge helps so much if you turn those dimensions around then it is
1x4^3/12 = 16/3 = 5.333. That's 4^3/4 = 4^2 = 16X times the stiffness
for only 4x the thickness.
Note typical E values for wood are roughly 1/10th that of common steel.
--