© 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,

**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

]]>© Jeff R. Filler, May 2022, Pell City

This is so fun I can hardly stand it. I’ll edit later. I am working on some NLT steps for my wife’s `Chicken Palace’. The steps will be `massive’, thus NLT. NLT stands for `Nail Laminated Timber’. Some day (soon) I’ll write an article on NLT for the modern small farm. For now let me just say that the idea of NLT is perfect for day-to-day building on a farm, void of heavy equipment (tractors, forklifts, and so on). The idea is this: if you want to make something that has massive, solid chunks of wood, but such chunks are unavailable, unaffordable, or unmanageable, then make them out of smaller (available, affordable, manageable) pieces, laminated (fastened) together.

Pictured, sticking out of the ground, are preservative treated (PT) 2×6 and 2×4 pieces of lumber. They are sticking out of the dirt of an old garbage dump on my property. The dump, and lumber, are decades old. How do I know? The garbage dump site was at one time a pasture. The pasture was left to nature and was swallowed up by Alabama jungle. From what I know about the history of the property, and the junk, it’s been there for decades. Several years ago I was cutting trails into the jungle, discovered the pile, and pulled from it PT lumber from which I made my wife an NLT bridge for her garden. The idea of using PT wood was driven by the fact that the garden area is outside, subject to the weather, plus the watering of the garden, and plus being a bridge of an intermittent `stream’ that becomes alive with most rainstorms. The idea of using PT lumber *from the pile* was motivated by: 1) I like to use materials on hand, whenever possible, and avoid a trip to the lumber yard; and, 2) if the lumber has already lasted several decades in the hot, humid, wet, dark, damp, Alabama jungle, it will certainly last a few more in whatever I use it for.

Granted, some of the lumber was indeed rotted, and disintegrating, but there were a surprising number of boards with appreciable integrity. Now, mind you, I’m not using this lumber to span great spaces; on the other hand, for any modest loads to which it might be put to resist, I can always proof test it, and indeed did, for the NLT bridge. (Yeah, I’ll write it up some time.)

Back to the story at hand. The step will be massive and fully bearing. If the board(s) survive excavation, they’ll probably be okay. In fact, I’ll lift on one end, putting some stress into them, for extraction, as kind of a preliminary test. Whoa! They both came out intact! I will next cut the 2×6 board to lengths suitable for the `lams’ (laminations) that will make up the step.

The beauty of this whole process is that I can carry the individual pieces (2 x 6 nominal x 30 inches actual) to the step location quite easily, and construct the step in place. And thus I accomplish the very important fact that I don’t have to carry some *whole* step, found or made somewhere else, to the place where I am going to use it.

(I’ll write up the NLT Bridge thing later.)

Okay, here it is! Don’t laugh, dammit! Four 2 x 6’s nail-laminated to a 6 x 6. All ancient PT pieces found (discarded) on the property. It only cost me 12 – 16d Sinkers, plus some kW-s of `electricity’ for my cordless saw. But what it saved me was my back! It’s just my wife and I on the farm at present, and I would rather make something in place, than have to get or make it someplace else, carry it. AND MY WIFE LOVES IT!!!

NOTES:

1) It is beyond remarkable that the lumber has survived all this time, relatively intact, with portions buried and above ground. I don’t doubt that there might be some strength loss, but the lumber is intact, and just looks `weathered’.

2) The 2 x 4 still has the remains of an end-tag. I can barely make out `CCA-C ABOVE GROUND’. It apparently did pretty good in contact, and below ground, as well. I wonder if I can get enough info off the tag to date it? I don’t think so; it’s partial, and faded, and the staples to attach it quite rusted.

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(c) Jeff R. Filler, May 2022, Pell City, AL

I haven’t tried one out yet … the Simpson Strong-Tie Strong-Drive® SDPW Deflector Screw … but the idea is fascinating … kudos to Simpson Strong-Tie. The description of the connector says that it is used to connect non-load-bearing walls to trusses and joists while resisting code-prescribed lateral loads. Importantly, the connector (presumably) does so while allowing the (trusses or joists) to deflect downward, under service loads. Indeed, they are able to accommodate up to 1.5 inches of differential downward displacement.

Source: Simpson Strong-Tie

The situation is this … one year we had particularly heavy snowfall in the town where I lived (and `Practiced’). In fact, the actual snow loads on roofs were reaching the then design snow loads for that jurisdiction. I was called one day by a concerned occupant/owner of a townhouse structure to look at the deflections in the ceiling above a great room. The deflections in the ceiling were indeed `noticeable’. The ceiling was (gypsum) attached to the bottoms of metal-plate-connected wood trusses. I did some measurements. The deflections were about an inch and a half. Let’s see. Code-limited deflections, per IBC 1604, are (were) `L/240’. That means that the roof trusses are `allowed’ (by Code) to deflect (sag) downward an amount equal to the span divided by 240, under design load. So, for a span of 32 feet (384 inches), the trusses are allowed sag 384 divided by 240 or 1.6 inches. Whoa! Nice! Though not for the owner. It appeared that the trusses were designed `right to the wire’ in terms of deflection. So the trusses themselves were okay (as long as it didn’t keep snowing). One could kinda see the `sag’ of the trusses overall, across the 32-foot span, but what one could really see, was where the trusses were resting on interior walls. Those trusses hardly deflected at all, and the differences between the deflections of the free-spanning trusses, and ones that didn’t, just 24 inches apart, were very easy to see, and causing problems! Interestingly, stepping outside the structure, and looking back at the roof from across the street, it was easy to see, as a sort of `print’ in the snow, where all the interior (non-load-bearing) walls were in the living space below.

It’s my suspicion that these SDPW deflector screws would address this situation, nicely!

Learn more about the SDPW deflectors screws here: www.strongtie.com/strongdrive_interiorwoodscrews/sdpw_screw/p/strong-drive-sdpw-deflector-screw

Endnotes:

1) At that time I was asked to visit several structures. Upon arrival in another I found the occupants seated for a meal in the dining area of a great room with similar free span. The sag in the ceiling above was about six inches. Yes, that was a problem. The excess deflection indicated failure of the roof system and possible soon catastrophe. Without going into detail on the construction problem (cause), I advised they vacate the room, immediately. Again, interestingly, stepping outside, walking up the hill of the back yard, and looking back over the roof, it was clear to see, `imprinted’ in the snow, were the locations of `non-load-bearing’ walls.

2) In such situations where the trusses or joists are connected directly to non-load-bearing walls, with no capacity for free vertical movement, the walls necessarily become load-bearing.

3) And, if 2) is true, then the floor supporting the wall necessarily takes on some roof load. (Let’s hope the truss isn’t a girder truss.)

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The deflection of a beam (plank, board, joist, etc.) due to the action of a concentrated load at midspan is …

Δ = P L^{3} / (48 E I)

where

P = the load (say a person)

L = beam (plank, etc.) span (distance between supports)

E = Modulus of Elasticity (E, `Eeee’, MOE)

I = Area Moment of Inertia of the `beam’ = b d^{ 3} / 12.

For `P’ … I used myself, clothed, and carrying, my cell phone (pictured), … 192 lb;

L = see each plank;

48 … a number based on beam curvature theory, the only number in this whole exercise that is `exact’;

E = see below; and

I = see each plank.

For E … I selected:

For the SP pieces, which, incidentally, were cut from the same log, merely inches from each other … E presumed = 1,300,000 psi (NDS *Supplement*, No. 3 and Stud; there were big knots!)

For the ERC … E presumed = 880,000 psi (*Wood Handbook*, MC 12%, beautiful wood, with only a few tiny knots)

Here is the specific plank data, calculated (theoretical) deflections, measured deflections, and back-calculated (actual) E values.

__Board Number ONE__

Southern Pine (1 x 5)

P = 192 lb

b = 4.94 in.

d = 0.844 in.

L = 60 in. (5 feet)

E presumed = 1,300,000 psi

Δ calculated (based on presumed E) = **2.69 in.**

Δ actual (measured) = **2.81 in.**

E actual (from actual Δ) = **1,243,000 psi **

(E actual is **4%** low of presumed)

__Board Number TWO__

Southern Pine (1½ x 8)

P = 192 lb

b = 8.0 in.

d = 1.56 in.

L = 84 in. (7 feet)

E presumed = 1,300,000 psi

Δ calculated (based on presumed E) = **0.717**** in.**

Δ actual = **0.75**** in.**

E actual (from actual Δ) = **1,243,000 psi **

(E actual is **4%** low of presumed)

__Board Number THREE__

Eastern Red Cedar (1 x 4)

P = 192 lb

b = 3.94 in.

d = 0.875 in.

L = 36 in. (3 feet)

E presumed = 880,000 psi

Δ calculated (based on presumed E) = **0.965**** in.**

Δ actual = **0.875**** in.**

E actual (from actual Δ) = **970****,000 psi **

(E actual is **10****%** HIGH of presumed)

__D____iscussion__

That the actual E’s for the Pine pieces turned out exactly the same (to 5 significant figures) is remarkable (1,242,951 psi and 1,242, 990 psi), particularly since nothing else was measured (is even *measurable*) to that accuracy.

That the SP actual E’s were low of published did not surprise me; I was a bit surprised they were as close as they were!

That the ERC E actual was higher did not surprise me; the Cedar was indeed beautiful wood, with few small knots.

One might argue (correctly) that I should be using `shear-free’ E for my planks. Indeed, the L/d ratios for these tests was LARGE. For large L/d the effects of shear contrastingly go away. Per NDS Appendix F, I could back out an approximate shear component deflection by increasing the published values by 3%. In that case the SP actual E’s would be more like 7 or 8% low of published, and the ERC E would be closer … to within about 7%.

One more thing: the Moisture Content (MC) of the ERC was 5-ish%; the SP 11-15%. MC has a role to play in E, and other structural properties of wood. Did that have a role in the actual SP E’s being lower and presumed, and the E actual for ERC being higher? Quite perhaps! Does it *matter*?

Finally, what about dead load deflection? The planks had already deflected due to their own weights (not much), before I climbed on, and set up `zero’ on the `gauges’. Thus, the measured deflections were due to `P’ only.

]]>(c) Jeff R. Filler, Pell City, 2022

Continuing our discussion on shear deflection of wood beams (joists, etc.), let’s look at a prefabricated wood I-joists. Basic design information for prefabricated wood I-joists is found in Chapter 7 of the *National Design Specification®* *for Wood Construction* (NDS). For more detailed design information Section 7 requires we go to the manufacturer or product code evaluation report. Let’s consider the APA PRI-40 9 ½” joist in ICC code evaluation report ESR-1405 (__link__).

And let’s consider the joist used to span 13.5 ft of floor, joists spacing (s) of 16 in. o.c., and carrying a floor `area’ load (σ) consisting of 40 psf Live (Occupancy) and 15 psf (Dead). For now let’s assume the Dead load includes the weights of the joists. The joists span simply between supports.

ESR-1405 says that the deflection calculations must consider shear deflection, and provides the following equation (for uniform loads):

δ = (5 ω l ^{4} / 384 EI’) + (ω l ^{2} / K’).

The first term we are familiar with … it’s the standard equation for deflection of a beam using beam deflection theory (flexural stress and strain). EI values for the joists are provided in the ESR. The (’) denotes that the value of EI must be adjusted for end use conditions, e.g., moisture and/or temperature, as required by Section 7.3 of the NDS.

The second term addresses shear deflection. K is a `shear stiffness coefficient’, also provided in the ESR, and must be adjusted, as required, by Section 7.3 of the NDS.

For this example we will assume that the joists are going to be used in dry and `normal-temperature’ conditions, so the adjustments on EI and K will not be necessary (adjustment factors 1.0).

For the PRI-40 9½ joist we read, from ESR-1405, EI = 184,000,000 psi ( = EI’), and K = 4,940,000 lb ( = K’).

The Live load on each joist is … w = σ x s … 40 psf x 16/12 ft = 53.3 plf … or 4.44 lb/in. (pli).

The Dead load on each joist is 15 x 16/12 = 20 plf, or 1.67 pli.

The Total load on each joist is 6.11 pli.

Getting L into inches, so our units work out … L = 13.5 ft x 12 in. / ft = 162 in.

Let’s calculate the deflection due to Live load …

δ = [(5)(4.44 lb/in.)(l63 in.) ^{4}] / [(384)(184,000,000 lb/in.^{2})] + [(4.44 lb/in.)(162 in.) ^{2}/(4,940,000 lb)]

= 0.217 in. + 0.024 in.

= 0.240 in.

Note that the shear term is 10% of the total. For sawn lumber we’re told that shear deflection is 3% of the total, and for glulam, 5%. I will leave it to your ponder as to why prefab wood I-joists are different.

If you put all this stuff in a spreadsheet, you’ll see that, with the equations above, the ratio of the shear deflection to total deflection increases with shorter span (and decreases with larger). While this might be instant cause for alarm, the deflections *themselves *are less for shorter spans, even relatively less. For short spans deflections rarely control!

The ratio of span to Live load deflection in the example above is … 162 in. / 0.240 in. = 674; 0.24 in. (about 1/4th of an inch), and δ (live) / L = 1 / 674 should be suitable for all but the most brittle floors.

]]>(c) Jeff R. Filler, Pell City, Alabama, 2022

But First, E apparent and E true

`E min’ is calculated wood property used in the design of structural wood members (beams, columns, etc.). It is a derived property, derived from the Modulus of Elasticity, E. The Modulus of Elasticity, E, is generally used to determine deflections of wood members such as beams, joists, rafters, girders. Values of E for sawn lumber, glued laminated timber (glulam), and other structural wood members can be found in the *Supplement* to the *National Design Standard*® *for Wood Construction *(NDS). Values of E for proprietary wood products, such as prefabricated wood I-joists and structural composite lumber (SCL), may be obtained from the product manufacturer and/or code evaluation reports. Predominately, E values are used directly in `bending’ applications; e.g., used for the calculations of deflections. These E values are used in `single-term’ deflection equations derived from `beam deflection’ theory. The equations address the strains generated by tension and compression stresses generated from bending moments acting in the beam, and, in theory, relate bending stresses to beam deformation (curvature) though the Young’s Modulus. In reality, however, wood beams deflect *a bit more* than as would be predicted using Young’s Modulus and the `singe-term’ beam-deflection-theory equations, due to the additional distortion and deflection caused by shear stresses in the beam. The additional deflection can be handled two ways. The first way is to include a `second term’ in the deflection formula which directly addresses deflection due to shear (`shear deflection’). The second way is to stick with the `single-term’ (beam-deflection-theory) equation and determine (in the case of testing), or use (in the case of design), an *apparent* Moment of Inertia that directly relates the total deflection to specific beam section, load, and support conditions. This so-called `apparent Moment of Inertia’ (E app, E apparent, and also often just *E*) is a bit *less* than the Young’s Modulus, or `true’ Moment of Inertia (and is, as said, reflected in the bit *more* deflection, due to the shear effect). In the case of sawn lumber, for example, values of E published in the *National Design Standard®* *for Wood Construction* (NDS) *Supplement *are to be increased by 3% (multiplied by 1.03) to obtain `shear free’ values (NDS Appendix F).

E min is derived from the `shear free’ E.

So, on our road to E min, we need to either determine, or obtain, `shear free’ E, a.k.a., true E. We can get it indirectly by calculation from the published E values that include shear effects, as described above, or, in other cases, directly, as published by wood products manufacturers, in the NDS *Supplement*, etc.

Once we have `shear free’ E, we can get E min.

Wood is a highly variable material. Wood properties vary between species, and among species. In the structural wood world, species, or species groups, are further subdivided into `Grades’. This `sorting’ into grades helps enables more efficient use of wood, as wells as helps pin down, or at least `narrow’ down, the range of properties with the Grades. But, at the end of the day, there’s variability in the properties of the woods within each grade.

For deflection calculations (using E app or E true) the variability is, for the most part, accepted as is. The published values for E are `average’ values for the species and grade. Some pieces will have more `E’, some less. As such, all other things equal, some members (beams, joists, etc.) will have a more, or less, deflection. And, again, this is generally accepted. Deflections are `serviceability’ issues … typically not *safety* issues. A little bit more deflection … `no big deal’. A bit *less* deflection, better yet! Where deflections are critical, a more stringent deflection limit could be set (to capture and corral some variability), or products with more precise properties could be used.

E min, on the other hand, is used for *stability* calculations, which are very much a *safety* issue. Stability issues for wood members include potential buckling of wood columns, posts, or studs, as well as lateral torsional buckling of beams (joists, rafters, etc.). Buckling of wood members may result in catastrophic rupture or collapse of the member itself, and even progressive collapse in a structural system. Thus while we might be somewhat `care free’ with the `E’ used in deflection calculations, we better be pretty darn tight with the E we use in stability calculations. Alas, E min.

The first step in getting E min (after getting E true), is to obtain E 05. This is the (supposed) value of E for which 95% of the members in that grade have equal or higher E.

In equation form,

E _{05} = E (true) (1 – 1.645 COV_{E})

where

COV_{E} = the coefficient of variation of E, representing one standard deviation relative to the mean standard deviation.

But that still isn’t good enough. Five percent of the wood members in that Grade could have lesser E, which could lead to catastrophe! We then take that number (E 05), and divide it by a factor of safety of 1.66.

Thus,

E min = E (true) (1 – 1.645 COV_{E}) / 1.66.

Selah …

Let’s say we’re dealing with sawn lumber, which has a published COV_{E} of 0.25.

And let’s say we’re starting with a published (reference) E (app) of 1,800,000 psi.

Then E true is 1,800,000 psi x 1.03 = 1,854,000 psi.

Then E 05 is E (true) [1 – 1.645 (0.25) ] = E (true) (1 – 0.411) = E (true) (0.589) = 1,092,000 psi,

and

E min = 1,092,000 / 1.66 = __658,000 psi__.

Nearly *one-third* of E (E app or E true).

I often wonder what this piece of lumber would look like. I think it represents a piece of wood with an E value between 2 and 3 standard deviations less than the mean … which, for a `normal distribution’ (if sawn lumber is `normal’) … represents, in this case, about one piece in 200. What would it look like? Mostly bark? Would it have a huge knot, or other big defect? Hopefully it would get `tossed’ (out) in the lumber sorting and grading process. If not, I would hope the `do-it-yourself-er’ at the lumber yard wouldn’t put it on his cart, or, if all else fails, the framing contractor wouldn’t dare use it for a beam or column.

]]>(c) Jeff R. Filler, 2022

A word on the accuracy of deflection calculations … wood beams, joists, etc.

In the example above (__link__) we calculated a deflection of 0.599 in. How accurate is this number? (I showed it to *three* significant figures … accurate to half a percent?) Leaving out, for now, the accuracy/precision of the other input, and the deflection equation itself, let’s look at Modulus of Elasticity, E. Take a look at Appendix F of the NDS (*National Design Specification®* *for Wood Construction*). It points out that the Modulus of Elasticity values in the NDS *Supplement* (which we used to calculate the deflection above) are *average* values. Yeah. I get that. Note that it goes on to say that individual pieces will have values higher and lower than the average. (Of course! … the age old panic that half the students in the class are performing `below average’!) And then the Appendix goes on to say how we might calculate the (lower) values. Let’s take a look!

Equation F-1, for example, can be used to calculate the E value at one standard deviation less than the mean, E _{16}, whereby we would get an E value for which we would expect 16 percent of the pieces (in that species and grade) to have a value less than that calculated. Let’s do it.

E _{16} = E (1 – 1.0 COV_{E})

For sawn lumber (from Appendix F), COV_{E} = 0.25, for glued laminated timber COV_{E} = 0.10, and so on.

So, for the (sawn lumber) example where E = 1,800,000 psi, E _{16} = 1,800,000 psi [1 – 1.0 (0.25) ] = 1,400,000 psi (0.75) = 1,350,000 psi. Whoa, 1 out of 6 pieces of lumber might very well have an E value of 25 percent of the average value, or even less … or … 1 out of 6 pieces of lumber might very well give us 25 percent *more* deflection than we calculated.

Note: the example linked above is that of a plank loaded (only) with a concentrated load at mid-span. It’s a real *life* scenario, but not one that would be a permanent part of a building. In an actual building we would not accept a deflection of 0.599 inches for an 8 foot span. We would come up with a different spanning member, or different framing system.

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E app (`Eee app’, `Eee apparent’, Apparent Modulus of Elasticity) is a design property for wood construction. It comes from the Modulus of Elasticity of the wood (E, MOE, Young’s Modulus). It’s not a `true’, or `pure’ property, and it is not the same as `True E’ (or True Modulus of Elasticity, or E true). It comes from bending beams to determine E, and it’s used in the design of beams (joists, rafters, girders, other spanning members). The background is this: beam deflection theory (an article itself) gives us equations (formulas) that we use to calculate the deflection (bend, bow, sag) of beams under load. Determining deflections is important as we don’t want beams to sag `too much’ … to where they might look, or feel unsafe, or otherwise cause problems. These equations are based on the flexural stresses in the wood. In a simple beam under downward load, the bottom of the beam is stretched, and the top of the beam is compressed, resulting in a curved or bowed downward shape. The equation used, for example, to determine the amount of deflection (`bow’) due to a concentrated load at the mid-span of a so-called `simple’ beam is …

Δ = P L^{3 } / (48 E I),

where

Δ = the amount of deflection, or sag,

P = the amount of load (e.g., the weight of a person, or downward load delivered, say, from post or column)

L = beam length

48 = a `number’, based on the distribution of the load and the support conditions of the beam (in this case the beam is supported at the ends, the ends are allowed to `rotate’, and the load is concentrated at the mid-span)

E = the Modulus of Elasticity, or Young’s Modulus (E), and

I = the Area Moment of Inertia of the beam shake (cross section).

Let’s take, for example, the deflection caused by a person weighing 185 pounds (lb) standing mid-span on a 2 x 12 Douglas-fir piece of dimension lumber laid flat (as a plank), supported at ends by saw horses 8 feet (ft) apart.

Thus,

P = 185 lb and L = 8 ft or 96 inches …

E = where to get? … Let’s see … we could look in our *Mechanics of Materials* textbook(s) from college, e.g., Gere, 5e, p 899 … shows, for Douglas fir … a *range* of 1,600 to 1,900 ksi. Hmmm, that’s not going to be very precise! (Of course, ksi stands for 1000s of psi, and psi stands for pounds per square inch.) Let’s try and be more exact. Wood is a highly variable building material. Wood is variable between species, and within species. To pin down E, let’s look to some resources more specific to wood. One source is the *Wood Handbook, Wood as an Engineering Material,* United States Department of Agriculture, Forest Service, Forest Products Laboratory … my hard copy says printed by the Forest Products Society, 1999. There’s a newer version available online (free). (I suggest you get it if you are at all interested in engineered wood! I consider it the Bible of engineered wood.) Page 4-12, Douglas-fir … we’ll pick `Inland West’ … 1.83 x 10^{6} psi. Let’s also cruise over to the 2018 American Wood Council (AWC) *National Design Specification*® *for Wood Construction *(NDS). Table 4A (in the NDS *Supplement*) gives us, for E, … 1,800,000 psi, for Grade No. 1 & BTR … 1,700,000 psi for No. 1, 1,600,000 psi for Grade No. 2, and so on. Which one do we pick? Pick the one that you intend to you, or already have on hand, or are willing to buy. (Or the one you end up requiring for your design situation.)

Back to I … the (Area) Moment of Inertia of the beam (plank) cross section.

For a rectangular shape,

I = b h^{3} / 12,

where

b = the beam (plank) width, and

h = the depth.

The actual dimensions of a so-called `2 x 12’ are 1.5 in. x 11.25 inches.

Since the 2 x 12 is laid flat, b becomes 11.25, and h = 1.5; so …

I = (11.25 in.) (1.5 in.)^{3} / 12 = 3.16 in.^{4}.

For 2 x 12 Douglas-fir Grade 1 & BTR (E = 1,800,000 psi),

Δ = P L^{3 } / (48 E I) = 185 lb (96 in.)^{3} / (48 x 1,800,000 psi x 3.16 in.^{4}) = __0.____599____ in__. (about 5/8^{th} of an inch).

I’ll bet we can *see* that! I’ll be you can also *feel* that if you happen to be the 185-lb load standing mid-span on the plank.

Now, we need to look at some fine print. If we look at the NDS, i.e., Appendix F, it says, first of all, that the `E’s found in Table 4A are *average* values, and that individual pieces will have higher and lower values. So, don’t be surprised if you actually conduct the plank scenario above, and get a value different than 0.599 in. In fact, you should *expect* a different value, more or less. Second, Appendix F3 tells us that the values include a `shear deflection component’. What is that? Well, when we bend wood, the wood fibers that are compressed *shorten*, the ones that are stretched *lengthen*, resulting in a curved shape, but the wood also `shears’ (rotates). Look at Gere p. 345 (stress) and p. 525 (strain). If we were to draw a perfect square on the side of the beam, it would shear (rotate) and become slightly `not square’, but a trapezoid. This `shearing’ causes the beam to deflect a bit more than just from the pure compressing and stretching of the wood fibers. If we read on, F3.1 says that the ratio of the shear-free E to the reference (table value) E is 1.03 (for sawn lumber). So, the published E values are ~ 3% *less* than true E (Young’s Modulus E, etc.), resulting in slightly *more* deflection (due to shear) than if we had used true E (Young’s Modulus) alone in the above equation.

If we backtrack, we’ll read on p. 4-2 of the *Wood Handbook*, that the value we got from 4-12 was from static *bending* and included (includes) the effect of shear. Further, it is stated that this effect can be approximately *removed* by increasing the table value by 10%. Footnote c to the table from which we plucked 1.83 (million psi) says the same thing. Backtracking further, we might have seen our first hint in all this in Gere, where, next to wood, for Modulus of Elasticity, it shows `(bending)’.One thing is certain: shear stresses in a wood beam* increase* the beam deflection. We could also say, with certainty, that an `apparent Modulus of Elasticity’, which is obtained by bending a beam, and back-calculating a modulus of elasticity using the `pure bending’ or `beam deflection formula(s)’, is *less* than the `true’ E (relationship between tension and compression forces, and resulting deformations). How did all this come about? Well, it’s more convenient, and relevant, to come up with a number for E for wood, using a bending test, versus some kind of compression or tension test. This is not necessarily a bad thing, either, since the huge bulk of engineering calculations performed on wood are indeed for its use in some kind of `bending’ application. (And accompanying shear stresses are essentially unavoidable in typical bending applications.) How much a beam actually bends (deflects, `bows’) is a big deal … an important design consideration. How much a wood *post* compresses under load, on the other hand, and based on a true E (without a shear effect), is rarely examined.

Several more things: 1) the current trend in the design of, especially, engineered wood products (EWP), e.g., prefabricated wood I-joists, glued-laminated timber, structural composite lumber (SCL), is to more deliberately separate (and perhaps publish both) the E app and E true values. We’ll talk more about this on another post; 2) formulas are available for explicitly determining the component of deflection due to shear stresses. The deflection equation thus takes on an additional term; the first term looks like the one we used above, accounts for the deflection due to elongation and shortening of the wood fibers, via Young’s Modulus, and uses the true E, and a second term that calculates the deflection due to shearing. Ironically, both components of the equation typically require true E for input. Theoretically, shear deformations/deflections are based on G, the modulus of rigidity, which is hard to pin down for wood. It is often taken as a function of (fraction of) E.

One thing that remains `not quite as certain’, at this point in this conversation, is the exact relationship (ratio) between `apparent’ E and `true’. Is E app 3% less (from the NDS, for sawn lumber)? … or E without shear 10% more (from the *Wood Handbook*)? Since the NDS is the `Standard’ used for wood design and construction, I’ll stick with the NDS value.… NDS Appendix F3 indicates ratios of shear-free E to reference E of 1.03 for sawn lumber, and 1.05 for glued laminated timber. What about SCL (structural composite lumber)? Prefab wood I-joists? We’ll examine these more in other posts. But, for now, all of this is almost becomes `academic’, contrast to the *variability* of wood, and its properties such as Modulus of Elasticity, which we’ll also discuss in another post.

There is another reason we might need E true, however, for a wood beam, and that’s for the determination of E min, which is derived from true E, and used for stability analysis. We’ll save E min for another post.

]]>(Steel for now; wood later)

E stands for Modulus of Elasticity, or Elastic Modulus, also Young’s Modulus. It is also abbreviated MOE (`Em – Oh – Ee’). By definition it is the amount of stress (tensile or compression) required to produce a unit deformation (elongation or shortening) of a material. Structural steel has an MOE of 29,000,000 psi (per inch per inch). This means that a stress of 29 MILLION psi (pounds per square inch) will stretch (or compress) a 1-inch (long) piece of steel … *1 inch.* Obviously, the piece of steel would have long before been destroyed. (You can’t stretch a 1-inch long piece of steel an additional inch before it breaks apart.) A more reasonable look might be 29 THOUSAND psi; a stress of 29,000 psi will stretch (or compress) a 1-inch long piece of steel … *1000th* of an inch. Most structural steel can `handle’ this kind of stress. And when we relax the steel (take away the stress), it will return to its original length. And that’s what we mean by `elastic’. We can push or pull on it, compressing or elongating it, and as long as we are in the `elastic’ range (of stress) … when we `let go’ it returns to it’s original condition (original length). When a material is `linear elastic’, the relationship between stress and deformation is `linear’ (straight line!); and we can say that deformation is *proportional* *to* stress (in the linear elastic range).

Let’s consider a stress of 29,000 psi, applied to piece (bar) of steel 100 inches long (about 8 feet). It will stretch … 1/1000th of an inch per inch … and (but) it’s 100 inches long, so, 1/1000th of an inch per inch, times 100 inches, equals … one-tenth of an inch! Small, but we could probably actually *see* it! It’s real! The beauty of steel is that most steels can handle that amount of stress, and, while it does deform, it’s not very much!

Here are the equations:

E = σ / ε …

where

E = Young’s modulus, or modulus of elasticity, or elastic modulus, E, MOE

σ = stress (force per area), and

ε = unit deformation, a.k.a., `strain’.

Strain, if you would like an equation, is …

ε = δ / L

where

δ = amount of deformation, and

L = length (original) of material being deformed.

Stress, if you want an equation, is …

σ = F / A

σ = stress,

F = force, either tension or compression

A = cross section area.

Now you can use the equations to confirm the number juggling above.

Let’s say we have a steel bar that is 1 inch x 1 inch (square) cross section, and 100 inches long.

If we apply a tension force of 29,000 pounds (a lot!) to both ends,

σ = F / A = 29,000 pounds / (1 inch x 1 inch) = 29,000 psi.

The strain will be … (rearranging the equation above)

ε = σ / E = 29,000 psi / 29,000,000 psi per inch per inch = 1/1000 inch per inch.

Then, (and rearranging),

δ = ε x L = 1/1000 inch per inch x 100 inches = 0.1 inch.

The 8-ft bar will elongate a bit less than 1/8th of an inch supposing the force is tensile. If the force is compressive, the bar will, likewise,* shorten* 0.1 inch, but only if we can keep it from buckling (like a noodle, like pushing on a rope)*.*

*A few more words on `Elastic’ … *

Let’s take, for example, Grade 50 ksi steel. First, ksi stands for `thousands’ of psi. And the 50 is the amount (of ksi) at which the steel *yields* (yield stress). Up to the yield stress (0 ksi to 50 ksi) the steel essentially elongates and compresses under the action of stress in accordance with `E’ (Young’s modulus). The steel deforms, but when the stress goes away, so does the deformation. Above the yield stress two important things happen. First, small amounts of additional stress cause (relatively) *large* amounts of additional deformation. We could also say that the relationship between stress and deformation is no longer linear (no longer governed by `E’). Second, when we take the stress away, the steel does *not* return to its original state (length). It gets a certain amount of *permanent* deformation. The steel behaves in a *plastic* fashion (no longer elastic). You can learn more about this via any (good) steel or engineering mechanics textbook.

But let me say a few more things, about *steel*. I have cast the conversation above in terms of a bar of steel (1 inch x 1 inch x 100 inches long), subject to a (tensile) force. If we apply a force of 1 pound (lb), we get a stress of 1 lb / (1 inch x 1 inch) or 1 psi; we get a strain of 1 psi /29,000,000 psi per inch per inch, and an elongation of 1/29,000,000 inch per inch times 100 inches or 1/290,000 of an inch (pretty small). A force of 100 lb would give us 100 times as much, and so on. Deformation goes with force (up to the yield stress, or elastic limit).

But now let’s think in terms of taking the bar and stretching it determined amounts of *deformation* (instead of applying force or *stress*). A tiny deformation will require a tiny force; a deformation ten times as much will require a force of ten times as much, and so on. Up to the yield stress. The deformation associated with a stress of 50 ksi is …

ε = σ / E = 50,000 psi / 29,000,000 psi per inch per inch = 0.0017 inch per inch …

and for the 100-inch-long rod …

100 inches x 0.0017 inch per inch = 0.17 inch (about 3/16 inch).

If we keep elongating the rod … up to 1/4 inch, 3/8 inch, and so on, we’ll continue to elongate the bar, but it will take on only a bit more stress (force). No matter how much more we stretch the bar, it will not take on (`carry’) much more than 50,000 psi, or, in this case, 50,000 pounds. Eventually the steel will … BREAK APART! The stress at which it (ultimately) breaks is called the `ultimate stress’. The deformation at which the steel ultimately breaks may be multiple times that associated with the yield stress. The ability of steel to take on a lot of deformation, before finally `coming apart’, lends to it being a `safe’ construction material, as well as strong. The steel will take on gigantic deformations, but still hold (things) together.

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