# 6 x 6 sawn post vs 3-ply 2 x 6

6 x 6 wood post versus 3-ply 2 x 6

© Jeff R. Filler, May 2022, Pell City

(DRAFT)

If you are an a professional engineer providing designs (or fixes) for wood-frame construction, sooner or later you will be asked, by the contractor / framer, if a 3-ply 2 x 6 can be used in substitution for a solid 6 x 6. Probable motivation of the question is that, one, pieces of 2 x 6 are generally more `available’ at the job site or lumber yard, and, two, 2 x 6s are easier to haul around, than 6 x 6s. So, let’s compare the `strengths’ of the two options, so we’re ready should we be asked the question. And we can do similar for 4 x 4, 8 x 8, and other size posts. (And if you don’t think an 8 x 8 solid wood post is heavy, go get one and haul it around.)

Let’s consider, for our illustration, Douglas-Fir/Larch (DF/L) sawn lumber, grade No. 2. And let’s consider a post height (either 6 x 6, or 3-ply 2 x 6), of 8 feet. Depending on the details of actual framing, such as for the post being framed within a wall for an 8-foot ceiling, the length (height) of the post will be different, say 92 ^{5}/_{8} inches. (I’ll let you look up as to why.)

**6 x 6 Post Unbraced**

Now let’s determine the allowable capacity (load) for the post. (I will use the terms `post’ and `column’ rather interchangeably in this `post’.) We’ll assume that the post is loaded centrically (down through the centerline of the post). We’ll assume that the post is not supported laterally from buckling, i.e., the post is out in the middle of a room, surrounded by `air’. The allowable capacity (allowable centric load) for the post, in accordance with the *National Design Specification®* *for Wood Construction *(NDS), is determined by multiplying the cross section area of the post, A, by the allowable compression stress, *F*_{c}^{ʹ}. And the allowable stress *F*_{c}^{ʹ} is the design value for compression parallel to grain, *F*_{c}, multiplied by all applicable adjustment factors addressing actual (end) (in service) conditions. For (let’s say) normal temperatures, dry (protected, indoor) conditions, the only adjustment factors we need to look at are the load duration factor, C_{D}, the size factor, C_{F}, (for softwood species other than Southern Pine), and the column stability factor, C_{P}. (Oh, and the wood is not incised, so C_{i} = 1.00.) For the load duration factor we’ll say that the post supports a `normal’ occupancy load, so C_{D} = 1.00. The size factor we get from the NDS *Supplement*, Table 4D, as 1.00. And the column stability factor, C_{P}, is, well, *more complicated*. Here is C_{P}, …

From NDS Section 3.7.1 …

C_{P} = [ 1 + ( *F*_{c}_{E}* **/ **F*_{c}^{*} ) ] / 2c + √ { [ 1 + ( *F*_{c}_{E}* **/ **F*_{c}^{*} ) ^{2} ] / 2 c – ( *F*_{c}_{E}* **/ **F*_{c}^{*} ) / c },

where

*F*_{c}_{E}* *= 0.822 E_{min}^{ʹ }/ (*Ɩ*_{e} / d ) ^{2}_{, }

where

E_{min}^{ʹ}* *is the adjusted `minimum’ modulus of elasticity used for buckling (stability),

*Ɩ*_{e} is the effective column height (length),

d is the depth of the post in the plane of (parallel to) buckling, and

*F*_{c*}* *is the compression parallel to grain design value multiplied by all applicable adjustment factors except C_{p} (good thing, since we don’t have C_{p}; otherwise the calc would be iterative!).

Crunching for C_{p} we get, along the way,

E_{minʹ}* **= *E_{min }C_{M} C_{t} C_{i} C_{T} = 470,000 psi (1)(1)(1)(1) ( … all the C’s are, without further explanation, 1.00),

*Ɩ*_{e} is *Ɩ *is 8 feet or 96 inches (post modeled as pinned top and bottom),

d is 5.5 inches (actual depth or width of 6 x 6 nominal dimension post), …

*F*_{cE }* *becomes 1268 psi,

C_{p} becomes 0.850,

*F*_{cʹ}* **= **F*_{cʹ}* *(C_{D})(C_{M})(C_{t})(C_{F})(C_{i})(C_{p}) *= *700 psi (1)(1)(1)(1)(1)(0.850) = 595 psi.

The corresponding allowable load on the post is

P = *F*_{cʹ}* *A = *F*_{cʹ}* *(b x d) = 595 psi (5.5 in. x 5.5 in.), giving

**P _{allow} = 17,995 lb … say 18,000 lb.**

That’s a lot of load! … (if loaded centrically … we’ll talk eccentric loads in another post).

Remember, in this (the above) scenario the post is `out in the open’ … surrounded by `air’ on all sides … it could potentially buckle in any direction. Because the cross section is square, the calculation above holds for both directions.

And note that the C_{p} = 0.85 shows us that the capacity of the post is reduced by 15% to keep it from `buckling’. If the post were really short, say 1 foot tall, or if the post were, say, embedded in the corner of, and held in place, by framed walls, it could not buckle, C_{p} would be 1.00, and the allowable (centric) load would be 700 psi (5.5 in. x 5.5 in.) = 21,175 lb (say 21,000 lb).

**3-ply 2 x 6, Unbraced**

Now let’s look at using 2 x 6s instead, and again the post is `out in the open’, such as supporting a beam above (out in an open space). And let’s start with 3 plies. Each ply has a cross section of 1.5 in. x 5.5 in. or 8.25 square inches. Three of them gives us 3 x 8.25 = 24.75 square inches. The 6 x 6 (5.5 in. x 5.5 in.) is/was 30.25 square inches. If we use `the same’ `Douglas-Fir/Larch No. 2’ we might think that we’re going to automatically get less capacity, based on cross section area. But, a No. 2 2 x 6, which is `dimension lumber’, will have a different set of design values than a 6 x 6 `post’. Design values for 2 x 6 dimension lumber come from the NDS *Supplement* Table 4A (not 4D, as for the post). For Douglas-Fir/Larch No. 2 we get, for *F*_{c}, 1350 psi, and, while we’re there, for E_{min}, 580,000 psi. Stated differently, you don’t get a No. 2 2 x 6 by ripping a No. 2 6 x 6. (Once you start ripping lumber, it needs to be re-graded!)

Let’s look at the brute compressive strength (allowable compression load) for the 3-ply 2 x 6 No. 2 Douglas-Fir/Larch, and by brute strength I mean it’s either really short, or it’s braced so it can’t buckle in any direction (C_{p }= 1.00). We’ll use the same conditions, i.e., dry service, normal temperature, thus C_{M} and C_{t} will both be 1.00. The size factor, C_{F}, from Table 4A, for 2 x 6 in compression, is 1.1. For this `supposed’ short-column / fully braced condition, the allowable compressive stress is,

*F*_{cʹ}* **= **F*_{cʹ}* *(C_{D})(C_{M})(C_{t})(C_{F})(C_{i})(C_{p}) *= *1350 psi (1)(1)(1)(1.1)(1)(1) = 1485 psi, and

The allowable compressive load is, for each 2 x 6 ply,

P _{allow, each 2 x 6} = *F*_{c}^{ʹ}* *A *= *1485 psi x 8.25 square inches = 12,250 lb.

For 3 of them,

P _{allow, 3-ply 2 x 6, short} = 3 x 12,250 lb = 36,750 lb.

This value holds for both directions (of potential buckling), supposing the post is so short (or so well braced) that buckling is not an issue (C_{p} = 1.00).

So, at this point, for a post that is really short, or fully braced against buckling, the 3-ply 2 x 6 is stronger, much stronger, than the 6 x 6, even though it’s `less wood’.

Now get back to the look at this post out in the middle of a room, 8 feet tall, unsupported against buckling. We’ll follow NDS Chapter 15 for built-up columns. First let’s look at the potential buckling in the direction of (parallel to) the wide faces of the 2 x 6s. In this case K_{f} = 1.00; I’ll let you read the fine print, and contemplate; it essentially equivalent to calculating the C_{p} (and allowable load) for a single 2 x 6 potentially buckling in the strong direction (of the 2 x 6) … (multiplied by the number of 2 x 6s). For a single 2 x 6 … *F*_{c}^{*} becomes 1485 psi; Cp = 0.709; *F*_{c}^{ʹ}* *= 1053 psi, and …

P _{allow, each 2 x 6, }_{strong direction} = 1053 psi (1.5 in. x 5.5 in.) = 8686 lb, or, for the whole post,

P _{allowable, }_{3-ply 2 x 6, }_{strong direction} = 8686 lb x 3 = 26,100 lb (rounded a bit).

We get better than the allowable load for the 6 x 6, 18,000 lb, free to buckle along the 8 foot height. Alternately, we could calculate the allowable load based on the whole section (3 x 1.5 x 5.5), and we’d get the same thing (buckling in the strong direction of the plies, centric load).

But we’re not done. Now let’s look at the 3-ply 2 x 6 buckling in the weak direction of the 2 x 6s.

We’ll consider the case of the plies being nailed (together) in accordance with NDS 15.3.3. (I’ll let you look at the details).

*F*_{c}^{*} is still 1485 psi; but C_{p} = 0.56; K_{f} = 0.60 (NDS 15.3.2), *F*_{c}^{ʹ}* *= 337 psi, and the allowable load on the 3-ply column is,

P 3-ply 2 x 6, weak direction = 337 psi (3 x 1.5 in. x 5.5 in.) = 12,400 lb (rounded a bit).

So, in the weak direction of the 2 x 6s we take a beating … 12.4k vs. 26k vs. 18k, and so on.

The allowable capacity of the built-up column is the lesser of the values for each direction, thus,

**P allow = 337 psi (3 x 1.5 in. x 5.5 in.) = 12,400 lb.**

If we add another lam …

P _{allow 4-ply 2 x 6} = 12,400 x 4 / 3 = 16,500 lb.

If we add yet another …

P _{allow 5-ply 2 x 6} = 16,500 x 5 / 4 = 20,400 lb.

We’re back in the 20,000-lb range for allowable load, but it’s a lot of wood, and will look weird (5-ply 2 x 6 with gross dimension 7.5 in. x 5.5 in.). Let’s *not* use the `n-ply’ 2 x 6 post out in the middle of a room.

Let’s look at a case where the column is *not* nailed in accordance with 15.3.3.

In this case we must look at the buckling capacity of each lam, in the *weak* direction of each lam. For C_{p} (without K_{f}) we get 0.077 (yikes), and *F*_{c}^{ʹ}* *= 144 psi …

P allow, each lam = … wait! … we get *Ɩ*_{e} / d = 64 … greater than 50 … *not even allowed*!

Whoa, do you mean to tell me that in no case is a framing member with 2 inch nominal thickness * (not) even allowed* in a condition where it is unbraced in the weak direction for a height of 8 feet?!!! … Yup!!! (We’ll, that’s not entirely true … we can go with a ratio up to 75,

*during construction*… see NDS. But that’s only for during construction.)

**3 – 2 x 6 studs embedded in wall, braced by attachment of sheathing**

Let’s look at a final case where the 3-ply 2 x 6 post is embedded in a wall. If the plies are nailed together in accordance with NDS 15.3.3, then the allowable load is the lesser of the 26k buckling in the strong direction (out of the plane of the wall), and 12k (in plane) … GIVING 12k! We don’t get any automatic benefit by `fitting it’ in the wall. We could add more lams, as needed, to get the needed `P’ (but, before doing so, look below). And, though it’s more wood, we wouldn’t `see’ it, since it’s embedded in the wall. BUT!

Better yet, let’s NOT call it a built-up column; instead, look at the whole load splitting up (preferably equally) into each lam, and look at each lam. (I think this is what NDS C15.3.2 is getting at, in a not-so-clear way.) And let’s `nail the hell out’ of the sheathing, or wall board, to each lam, considering each lam like a `stud’ … preventing each lam (stud) from any movement (buckling) in the plane of the wall (weak direction). We could use an effective bucking length equal to the vertical nail spacing; let’s just say we have a LOT of nails, so that the effective buckling length is small, and C_{p} ~ 1.00. Using a nail spacing of, say, 6 inches, giving *Ɩ*_{e} = 0.5 feet,

*F*_{c}^{*} is still 1485 psi; but C_{p} = 0.99; *F*_{c}^{ʹ}* *= 1470 psi; P allow, each = 12,100 lb (rounded); 3 of them will give … 36,000 lb!

P _{allow 3-plies of 2 x 6, in wall }= 36,000 lb with respect to buckling in the plane of the wall …

Going back and checking buckling out of the wall … P = 36,750 lb (from above).

The controlling capacity is the lesser … 36,000 lb based on buckling out of the plane of the wall, versus 36,750 buckling in the plane of the wall …

**P allow = 36,000 lb!**

**Summary:**

6 x 6 Douglas-Fir/L No. 2 column, unbraced length 8 feet (out in the middle of a room) … 18,000 lb.

3-ply 2 x 6 DF/L built-up column, unbraced, 8 feet tall (out in the middle of a room) … 12,400 lb.

3-ply 2 x 6 DF/L built-up column, built into a wall, nailed in accord with NDS 15.3.3 … 12,400 lb.

3 – 2 x 6 studs, equally carrying a compression load, next to each other, each nailed to resist buckling in the plane of the wall … 36,000 lb.

6 x 6 DF/L post, embedded in a wall, nailed to resist buckling in the plane of the wall (but not out-of-plane) … 18,000 lb.

6 x 6 DF/L post, embedded in a wall *corner*, nailed/framed/sheathing to resist buckling in any direction … 21,000 lb.

**Conclusion**

For a sawn column out in open space, i.e., unbraced for its height (floor to ceiling), use solid sawn lumber (or engineered lumber). For a column embedded in a wall, use multiple individual members (`studs’) side-by-side, nailed tougher, and, importantly, each stud braced to prevent buckling in its weak direction by attachment of wall sheathing or board to the narrow faces. (Yeah, it’s a built-up column, but not technically.) Of course you can use a solid rectangular post in a wall, instead of a number of 2 x’s, if available, affordable, and meets the required load.

Sooner or later you will find, or make, a spreadsheet, to handle the column calcs. And, at some point in time, on some drawings, you may specify something like `6 x 6 DF/L No. 2 or 3-ply 2 x 6 DF/L No. 2’, provided both options do indeed work.

JRF