… and now for the headers

As an engineer I tend to design structures from the top down. First the roof loads, which determine rafters, or trusses, or whatever roof framing members. Then the beams/headers/girders supporting the rafters, then the columns or walls supporting the headers, and so on. Top-down load path. But that’s not the way the structure is built!

2×6 rafter resting (precariously) on 3-ply 2×6 header

Now for the headers …

Starting from the top … we found that we’re going to use 2×6 rough sawn Southern pine rafters … spanning 12 feet, and overhanging 1 foot each end. The design was `governed’ by a concentrated load (construction worker) at mid-span of the rafter. The 2×6 rafter also accommodates a uniform construction load of 20 psf, as well as 25 or so psf of wind uplift. The barn has two 12-ft x 12-ft bays, and is `pole-barn’ construction, meaning the gravity loads (and lateral loads) are resisted (carried down … to the ground) by the posts/columns. `Headers’ (beams/girders/girts … whatever you want to call them) will support the rafters and transfer the loads to the posts/columns.

Each header will need to carry one set of ends from the rafters. In addition to the rafters spanning 12 feet, and overhanging 1 foot each, I need to have some kind of rake, so I’ll assume the headers also span 12 feet (the other direction, capturing the ends of the rafters), and (also) overhang 1 foot. Otherwise I’ll need to come up with some kind of outrigger system.

Let’s start out with the headers simply spanning 12 feet between posts/columns. This will be slightly conservative with respect to `bending’, since it ignores the counter flexure from the rakes. I’ll assume the roof weighs 5 psf (relatively light). In terms of a uniform, or line load, the headers will need to carry (between the posts) …

Dead Load …

w DL = (½ of 12 ft1 + 1 ft overhang) x 5 psf = 7 ft x 5 psf = 35 plf …

or

W = w x l = 35 plf x 12 ft = 420 lb … (dead weight of roof on the header span).

Live Load …

A single construction worker would contribute 300 lb live load … (probably won’t govern) …

A construction crew/operation will deliver … 7 ft x 20 psf = 140 plf … or 140 plf x 12 ft or 1680 lb2. (Yup, the whole crew governs over a single worker.)

Let’s look at the total … roof weight plus construction crew = 420 lb + 1680 lb = 2100 lb total load, uniform.

Check … total uniform `area’ load … 5 + 20 = 25 psf.

Total load to each header = uniform load x area supported by each header = 25 psf x [(½ of 12 ft + 1 ft) x 12 ft] = 25 psf x 84 sf = 2100 lb. Check.

Assuming the design of the header is `controlled’ by bending (stress) … let’s find the minimum Section Modulus needed.

In equation form …

Let fb = applied stress under load = Fb’ = allowable stress, where

fb = M / S,

and

assuming an allowable bending stress, Fb’, of … say, 1000 psi …

M = bending moment in header under design (applied) load = W L / 8 …

W = 2100 lb;

L = 12 ft;

8 = 8 …

M = 2100 lb x 12 ft / 8 = 3150 lb-ft.

Let’s get this into lb-in. …

3150 lb-ft x 12 in./ft = 37,800 lb-in.

Doing some rearranging …

fb = Fb’ = M/S … gives S needed …

S needed = M / fb = M / Fb’ = 37,800 lb-in / 1000 psi = 37.8 in.^3.

What sawn lumber (Southern pine) sections give me 37.8 in.^3???

Looking the the National Design Specification ® for Wood Construction (NDS) Supplement …

2 x 12 dimension lumber (1.5 x 11.25) … S = 31.64 … not enough.

Try 2 x 12 `actual’ … S = b h^2/ 6 = 2 x 12^2 / 6 = 48 … ENOUGH!

I will ask my son-in-law to cut me a 14-ft 2 x 12 (actual dims) header!

WAIT!

Several problems.

First, my son-in-law can cut sawn lumber in dimensions (actual) up to 6 inches, easily … 12 inches, more difficultly.

Also, I have been carrying the 2 x 6 rafters around … 14 ft long … and they are about all I can handle by myself … they are remarkably heavy; I think it’s because they are `GREEN’.

What are my other options?

2 x 10 actual … S = 33.3 … not enough.

4 x 10 actual … 67 … way enough.

4 x 8 actual … 43.7 … enough!

But that is still going to be heavy …

The `line weight’ of a wood member goes like … w = γ x A, where γ = specific weight, and A = cross section area. I could compare line weights, but I don’t know the specific weights of this `green lumber’ here3 … so let’s compare cross section area …

2 x 6 = 12 square inches (actual) … (I can lift into place)

4 x 8 = 48 square inches … that’s four times as much … will be four times as heavy … no way! I’m not going to ask my son-in-law to come over with his tractor to lift each header … if I can avoid doing so.

Hmmmm.

Try 6 x 6 … S = 36 in.^3. Enough! Actually, not quite … BUT … I’ll bet I’ll be able to split hairs, and make it work.4

Turns out A = 36 in.^2. Same number; different property.

At least my son-in-law can more easily cut the 6 x 6.

Or I can laminate three 2 x 6s side by side.

But still way to heavy to heave up above my head.

OMG! I don’t have to nail laminate the header together on the ground … this is what I’ll do … I’ll lift each 2×6 ply into place, one at a time … something I can do … and nail laminate them together once they’re up in place!!!5

*****

This `three’ is very interesting. Remember that the rafters span 12 feet … and that the headers themselves span 12 feet … same support condition … 12 foot span. And same load condition, … roof dead plus live load. The rafters are spaced 2 feet apart … so each header carries the ends of 12 ft / 2 ft per rafter = 6 rafters. Since the rafters are supported by headers at each end … we could say that each header carries half of 6 rafters, or the same as three rafters! A 3-ply 2×6 header will carry the ends of three 2 x 6 headers.

Interesting.6

Selah

The rest is easy … especially if we look at it in the context of a 3-ply 2×6 carrying 3 – 2×6’s. Deflection should be good. Shear should be good. Bearing should be good, as long as each 2 x 6 ply of the header has the bearing required for each of the 2 x 6 rafters … or the header as a whole has enough bearing … and so far, we’re way good in bearing. Let’s look!

BEARING!

Bearing (rafter on header) …

The roof is sloped, so I’ll bevel one of the lams of the 2×6 header to match the slope of the rafter. That means that the rafters will only actually bear on one ply of the header … but we already determined that would be fine … we assumed that we would have bearing area of at least 2 in. (width of rafter) x 1.5 in. (code min) … but the beveled lam of the header is 2 inches wide, … we’ll have 2 in. (width of header) times 2 in. (width of lam) = 4 sq. in. … way good!

JRF

___________________________

1 … in the direction of the rafter.

2Note that we don’t combine the concentrated single construction worker and the uniform construction crew … it’s either or … or … as we crunch along down through the structure, the worst of the two. It (the concentrated load) governed the bending stress limit state for the rafter … but will not govern the bending stress limit for the header (the uniform load representing the crew, materials, etc.).

3… plus the green weight is changing day by day as the lumber dries out.

4At the very least, I’ll be able to employ the repetitive member factor, Cr, since I have 3 plies … Cr is (at least) 1.15 for this condition … 1.15 x 1000 = 1150 psi. S needed = 37,800 / 1150 = 32.9 in.^3. I’ll have 36. Yay! (We’ll talk more about `allowable stress’ later.)

5This (3-ply) solution is not the most conservative with regard to wood mass … but just think! … I can proceed without employing additional labor (son-in-law and his sons), as well as machinery (his tractor). I’m thrilled that I can proceed with `just’ 2x6s … at my disposal. Plus, I am surrounded by Southern pine …

6We could also look at it like this … each rafter carries 2 ft x 12 ft = 24 ft of roof, at 12-ft span … and each header carries ½ x 12 x 12 = 72 sf of roof (okay, not counting overhangs) … 72 = 3 x 24 … if one 2 x 6 carries 24 sf of roof at 12 ft span, then 3 of them can carry 72 sf at 12 ft span. (This only plays out with this particular framing layout.)

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