No, 4x12 is not common. Making up your own as suggested will make a far stronger and straighter header. When making it, try to pick fairly straight stock and if it is a faily long header, lay it out so the curves are opposite, i.e., if the top one curves left, the bottom one curves right. Start nailing from one end and keep pulling the other end together. That makes for a nice straight beam. You might (probably will) need clamps to draw the ends together when you get near. I have done that often with stock up to 2x10, haven't tried with
Hi. Is it common or uncommon for Home Depot and Lowe's to stock 4 x 12 lumber? I wanted to use a 4 x 12 for a header, but was unable to find it at HD, and didn't know if it was this particular store, if it has to be special ordered, etc. Thanks.
Depends on the wood I would guess. I couldn't find 4x12' Ceder at any of the home depot's or lowe's in the Metro Detroit area. I had to use 2
On Wed, 17 Aug 2005 08:04:59 GMT, "Eric and Megan Swope" scribbled this interesting note:
4X12 is a common size, but you won't find it at any Home Depot or Lowes I've ever frequented. To find sizes like this you have to find a real lumber yard, not the silly, poorly stocked and over priced "Home Centers."
I was recently in an 84 Lumber just up the street, which is a real lumber yard, and they had 4X12 rough cedar beams, as well as other sizes.
For a header, that isn't exposed so it doesn't matter what it looks like, I would laminate one up for myself. I would probably build it in place because it would be a bit heavy once assembled. Use 1/2" CDX plywood glued between two 2X12 boards and you will have a 4X12. And it will be stronger than a solid piece of lumber would be. Gang nail or bolt it together, or both. If I wanted this exposed, I would probably still build it like this and then dress it up on the outside with whatever grade wood I wanted.
-- John Willis snipped-for-privacy@airmail.net (Remove the Primes before e-mailing me)
similiar old building or a speciality lumberyard. does it have to be a 4x12? just nail two 2x12's together with some 1/2 ply sandwiched in the middle.
Thanks guys. I had considered that, and that is probably what I will do is use 2 x 12s with the plywood in between, just didn't know if 4 x 12s were commonly available.
Depending on where you live, you can often find rural sawmills where they can make any size timber a tree will allow. Of course they will be rough sawn, and you will have to rely on the type of wood they have, or supply your own tree.
I agree that nailing two 2X12's together is stronger than a single beam.
Why would that be? There's no more (and maybe less) material (neglecting the effect of an additional 1/2" (say) ply which would add some additional resistance.
On Wed, 17 Aug 2005 15:34:52 -0400, "CL (dnoyeB) Gilbert" wrote Re Re: 4 x 12:
No, the tube is *stiffer* than the solid rod of the same material. Our Strength of Materials teacher in 3nd year engineering went through the math/physics for the proof and I remember being stunned by it. But it was clearly correct as I recall.
I'm not sure what laws you're thinking of...let's see--if we consider a simple beam w/ uniform load w/ simple support at both ends the maximum deflection at the center is 5/384 (W*l^3)/(EI) where
E = modulus of elasticity (material property only) I = moment of inertia (dependent on geometry) W = applied load l = length
Now for a rod Irod = MR^2
where M = mass of beam and R = radius
and for a hollow tube it is Itube = M(R1^2 + R2^2) where
R1,R2 are inner/outer radii, respectively.
This superficially makes it look like Itube>Irod for R2 = R, but that doesn't include the mass which will be less for a hollow tube than for a solid rod.
Since we're after comparing two geometries of the same material, we can consider the density of the two to be the same as well as the length. On that basis, for the rod the weight/unit length is
mRrod = density*pi*R^2/4
and similarly,
mTube = density*pi*(R2^2-R1^2)
Substituting into the formulae for I the geometrical terms for each M we get that for each the moment of inertia is proportional to
iRod ~ R^4
and
iTube ~ (R2^2 - R1^2)*(R1^2 + R2^2) = R2^4 - R1^4
Thus, it can be seen that the moment of inertia for the tube section is always slightly smaller than that of the solid rod and since I is in the denominator of the deflection, the larger deflection will occur for the tube, not the rod for R2==R (the outer diameters equal).
If you figure on an equivalent weight basis, the tube will be stronger as the same amount material will be located at a farther distance from the neutral axis.
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