From the earlier post, amount of lumber is often described in terms of `board feet’. One board foot is `a board 1 inch thick, 12 inches wide, and 1 foot long’. Or equivalent. And the dimensions are nominal. A 1 x 12 that is 2 feet long has 2 board feet of lumber. A 2 x 6 that is one foot long has 1 BF. And so on. A 1 x 6 that is 2 feet long has 1 BF. Two 2 x 6’s side-by-side have 1 BF per foot of length. Two 2 x 6’s stacked on top of one another, 10 feet long, have 10 BF …

Cost of lumber is also often in terms of … per board foot, or per 1000’s of board feet (MBF).

….

Glulam is made up of laminations, the laminations of which are most often described in `nominal’ terms. For example, a Douglas fir glulam beam that is 5-1/8 in. x 12 in. x 16 feet long is made up of 8 (eight) 2 x 6 laminations, wide faces laminated (glued) together, each 2 x 6 (nominal) being 1-1/2 x 5-1/2 (actual), all 16 feet long. A `stack’ of eight 2 x 6’s. So, 8 x 1.5 = 12 … the depth of the glulam, and the width, 5-1/2, gets planed down to 5-1/8. How many board feet of lumber are in this beam? We look at the input lams. Each lam is 2 x 6, having 1 board foot per foot … there are 8 of them, and they are each 16 feet long …

Each glulam has … [ (2 x 6) / 12 ] x 8 x 16 … 128 BF.

10 of these beams … 1,280 BF.

100 of these beams … 12,800 BF … or 12.8 MBF.

Now go find the price of Douglas fir laminating stock, per MBF.

It will depend on the grade of laminating stock, and several grades go into a typical glulam beam.

]]>*“I don’t do countertops.”*

I tasked her with finding out how thick countertops are, and how many square feet she needs. I handed her a tape measure. “Tell me how many `square feet’ you need.” Then I headed outside to measure the tree.

I’d rather not fuss with the chainsaw mill, but her birthday is coming up.

A bit later she said, “35.68 square feet.” I said, ”whoa, nice!” (I saw that she started with measuring the counters in `inches’.) “ … I divided by 12 and came up with 404.16. But I knew that wasn’t right; a tiny house is 400 square feet. So then I divided by 12 again, and got 35.68.” “Good job!”

… when I was teaching, I would make my students do a “does the answer make sense?” check. “Good job!” She checked her answer by something she knew. Awesome. I would have just walked up next to the counter, paced off 15 or so feet, and multiplied by the width, 2-ish feet … 32 square feet.

… (but) 404 square feet of countertop would be absurd (in our kitchen) … I’d chastise my student hard for an answer like that.

“How thick?” She said … `5/4’. “One layer or two?” … `One’.

Okay, 5/4, as I remember, stands for 5 fourths of an inch. Oh, and it’s a `nominal’ dimension. (Nominal means, `in name’.) So, it’s not actually 5/4ths of an inch thick, but a bit less. For now, let’s say an inch, `actual’, or maybe 1-1/8 inches. The actual volume of wood that she needs is … 36 square feet times 1.125 inches, or 405 square feet-inches. But that’s a weird number (and, presumably only a coincidence nearly equal 404) … let’s divide by 12 (inches per foot): 405 square feet-inches / 12 inches per foot equals … 3.375 cubic feet … let’s say 4 cubic feet of wood.

Now let’s look at the log. The segment I want to use is 7 feet long, and diameter about 16 inches. When I take my mill to the log, I’ll be cutting off some of the `round’, so, let’s consider a square section that `fits’ inside a 16-inch diameter round (circle). (I’ll let you prove that the sides of the square, that just fits, are 1/√2 times the diameter … or 0.707 times 16 equals 11 inches. The square section has an area of 11 times 11 or 121 square inches, or 121 divided by 144 gives 0.86 square feet. The *volume *of (supposedly) available wood is 0.86 times 7 or 6 cubic feet. Six available versus four needed …

It* looks like* I’ll have enough wood, BUT, a couple things: first, the chainsaw (mill) turns a significant amount of wood into sawdust, and, second, having done this before, but for flooring, it seems I always end up milling way more wood than I thought I needed. We’ll see. In the mean time, I’m glad she didn’t say `2 inches’ … (thickness).

…

**Board Foot Measure**

If for some reason I don’t succeed at cutting up my neighbor’s tree for countertop, Linda has been looking into materials herself. Passing by the YouTube video she was watching, I heard something about 5/4, and `Board Feet’. Board Feet! … or Board Foot Measure! … I love it! (BF, BFM!) Lumber is often measured in `board feet’. I didn’t know that countertops are measured such, but, again, I don’t do countertops. Let’s calculate what she will need for countertop materiel, in board feet.

Board Foot Measure utilizes `nominal’ dimensions.

How many feet, `board feet’, or material will she need?

Linda measured 4850 square inches of countertop needed. I went over and measured 25-or-so inches wide. So, assuming she also used (about 25), she must have measured about 4850 / 25 or 1294 inches, or … 16 feet. Seems about right.

Then to the NDS `Supp’ (*National Design Specification® for Wood Construction, Supplement – Design Values for Wood Construction*). Table 1A provides sizes of boards. Boards are available in nominal sizes from 1 x 2 up to 1-1/2 by 16. Since nothing is 25 inches wide, let’s consider two pieces of, say, 1-1/4 x 14 (or maybe 16). These are `nominal’ sizes … nominal means `in name’. So, a 1-1/4 x 14 is not actually 1-1/4 inches by 14 inches. The Table also provides `Minimum Dressed’ dimensions. For the 1-1/4 x 14 it shows … 1 (dry) and 1-1/32 (green) by 13-1/4 (dry) and 13-1/2 (green) (inches). (This reminds me … the tree was living just a few weeks ago, thus `green’. After milling, I will need to let dry for a while, before installing. I hope she’s not in too much of a hurry for those counters.)

So, two pieces side-by-side (dry) is … 2 x 13.25 or 26.5 inches … should be good.

So, she could ask for 2 pieces of 1-1/4 by 14 x 16 feet long.

How many `board feet’ … if she’s asked.

“How many board feet of 1 x 16?”

Board foot measure … here’s how I remember: a one-foot-long 1 x 12 is ONE board foot.

We could say … board foot measure, BF, or BFM, is thickness (in inches) x width (inches) x length (feet), divided by 12.

BF = b x d x L / 12.

A 16-foot piece of 1 x 14 is … 1 x 14 x 16 / 12 = 18.67 board foot of 1 x 14 lumber.

Two pieces is 37.3 BF (of 1 x 14).

But before she marches off to the building supply store … these are countertops, and there is a corner, and there is cutting around the sink, and stove, etc. There’s going to be some waste, trim. Let’s say 15% … and by that I actually mean, `let’s order 15% extra, since there will be some waste’.

So, 37.3 + 0.15 x 37.3 = 42.9 … say 43 BF.

If, we use 1-1/4 x 16 instead … 43 BF x 16/14 = 49 BF. (I’ll let you check my math.)

…

Some more on BF, or BFM.

How may board feet are in a cubic foot of lumber?

One BF is 1 x 12 x 1 foot long … (one square foot x 1 inch thick).

If it’s 12 inches thick (one foot) instead of 1 inch thick … then it’s 1 x 12 BF.

We could say that BF measure is `cubic foot measure’ … multiplied by 12.

Or volume measure (in cubic inches) … divided by 144.

…..

BF repeated (reworded)

… to understand (or remember) `board feet’, think of a `board’, that is, say, 12 inches wide, and however many feet long. One board foot is a one-foot length of one-inch-thick board, twelve inches wide. If you have a board that is 2 inches thick, and 12 inches wide, and one foot long, it is … 2 board feet. More often, our boards are more than one foot long. Say the 1 x 12 is 2 feet long. It is 1 board foot per foot, so 2 feet will be 1 board foot per foot, times 2 feet, or 2 board feet (2 BF, or 2 BFM).

A 2 x 6 board, even though only half as wide, is twice as thick, giving us the same as 1 x 12. So a 2 x 6 is 1 BF per foot. A 2 x 12 (that is) 1 foot long is 2 BFM. A 2 x 12 that is 10 feet long is a total of …. 20 BFM.

If you want a formula … BF, or BFM = (b x h) / 12 … x L, where b is the thickness, in inches, h is the width, in inches, and L is the length, in feet.

So, an 8-foot long 6 x 6 is … [ ( 6 x 6 ) / 12 ] x 8 = 24 BF.

Maybe we could cast the formula as,

BF or BFM = [ ( b h L ) / 12 ] BF / in.2-ft.

Let’s try it … 1 x 12 x 4 feet long … 48 in.2 ft / 12 x BF / in.2-ft … or 4 BF.

… I’m making it too complicated. Just remember …

1 board foot is … 1 inch thick, by 12 inches wide, by one foot long, … and go with multiples (or fractions) from there.

And don’t forget … Board Foot is based on nominal dimensions.

KEY WORDS: nominal, dimension, dimensional, nominal dimension, BF, BFM, board foot, board feet, board foot measure, lumber, wood

]]>Buckling equations for wood members (beams and columns) depend on E. `Buckling’ is different than `bending’. When a wood member bends, it simply `flexes’, more or less, depending on the member, load, and support conditions. To make sure it doesn’t bend so much that it breaks, design values for bending, shear, etc., take into consideration variation in wood properties, and are based on `fifth percentile’ values, plus factors of safety. (Again, for another conversation.) This fifth percentile thing, and factors of safety, assure `breaking’ will *never happen*. The `fifth percentile’ thing means that the design values (for bending, shear, etc.) are based on properties well below the mean properties.

How much a beam bends (sags, deflection), on the other hand, before it breaks (though it never will), are based on average E. This is because deflection (sag) is a serviceability issue, not a `safety’ issue. The other calculations assure the beam (column, etc.) is `safe’; deflection calculations simply let us calculate the sag … ½ inch at midspan, ¾ inch, or whatever. The calculations, though they might make you think (by their complexity) they provide exact answers, … they do not. And how could they, if they are based on average values, but individual beams, joists, columns, etc., going into structures, will have values different than the mean values (some lower, some higher).1

Back to buckling. Buckling is different. Buckling is a `stability’ issue. Buckling of a wood member can be catastrophic (sudden collapse). For wood members in building design the two main buckling issues are lateral-torsional-buckling of beams (joists, rafters, etc.) and Euler buckling of columns (posts). The equations used to investigate buckling utilize E. Since buckling can be catastrophic, and since we actually want buckling to never occur, we don’t want to use just any average E, especially knowing E can be so variable. We use Emin.

Emin is determined by finding, again, the 5th percentile E, and dividing that value by a factor of safety of 1.66.

We can see this in, for example, the formula for Emin, in Appendix D of the National Design Specification® for Wood Construction (NDS).

Emin = E [1 – 1.645 COVE ] (1.03) / 1.66,

where,

the [1 – 1.645 COVE ] gets us down to the 5% level on E,

1.03 adjusts from apparent E to shear-free or true bending E (another conversation), and

the 1.66 is a factor of safety.

Selah.

Back to this idea of `never’. Let’s look at an example … a 24F-V3 Southern pine glued laminated timber beam, and the NDS Supplement Table 5A Expanded, as well as Appendices D and F in the NDS. From the NDS Supp 5A we get Ex true = 1,800,000 psi; Ex app = 1,800,000 psi, and Ex min 950,000 psi.

Following NDS App D we can see that Ex app x 1.05 = 1,800,000 x 1.05 = 1,890,000 psi, round to 1,900,000 psi = Ex true (shear-free) … yeah!

Then, Ex true [1 – 1.645 COVE] = 1,890,000 psi [1 – 1.645 (0.10)] = 1,580,000 psi … where 0.10 is the COVE for glued laminated timber (NDS App F). This is the `E05’, or `fifth percentile value’. Not more than one in twenty beams coming off the assembly line will (theoretically) have a true E (Ex) of less than 1,580,000 psi.

(This 1.645 is `1.645’ standard deviations below the mean.)

Now let’s divide by 1.66.

Ex05 / 1.66 = 1,580,000 psi / 1.66 = 950,000 psi … yeah, as from NDS Supp 5A.

Let’s calculate Z.

Z = (x – μ) / σ , where `x’ is 950,000 …

σ = standard deviation = COV times the mean, μ, … 1,890,000 x 0.10 = 189,000 psi.

So, Z = (950,000 – 1,890,000) / (189,000) = -4.97

We’re 5 standard deviations from the mean.

My Z calculator2 shows … 0.0 probability.

Let’s try another Z calculator … oh, 3.4 x 10-7 …

One in 3 million of these 24F-V3 beams will have an actual Ex of (less than) 950,000 psi.

Almost `never’.

I have often wondered what this beam, or board, would `look like’? Let’s do something similar for sawn lumber. Let’s do visually graded Douglas fir No. 2 … NDS Supp Table 4A gives E = 1,600,000 psi. This is an apparent E. So, per NDS App D, E true = 1,600,000 x 1.03 = 1,650,000 psi. Then to E05 … 1,650,000 [1 – 1.645 (0.25)] = 970,000 psi … noting the high COVE of 0.25, from NDS App F.

Then we divide by 1.66 … getting Emin = 585,000 psi … yeah, agrees with Supp Table 4A. Let’s calc the standard deviation … σ = 0.25 x 1,650,000 = 412,000 psi. Now for Z …

Z = (585,000 – 1,650,000) / 412,000 = -2.58.

The corresponding probability that we’ll run into a board with E (actual) less than 585,000 psi is … from a Z calculator … 0.00494 … or 0.005!

About 1 in 200 boards will have E true of less than 585,000 psi.

I have often wondered … what would this board even look like? The board has only about a third of the stiffness of the average for the grade and species. Is it falling apart? Rotted? A huge knot? Or wane? Some other huge defect? Would it even make it onto the grading table? Maybe? Would it make it off the other end?

So, and interestingly, you’re more likely to run into a sawn piece of lumber that has an actual E equal to or less than Emin, contrast to, say, a more-controlled piece of lumber, like glulam … because of the COVE. So, while we would `never’ run into a glulam beam with actual E less than Emin, we *might* with sawn lumber.

Here are the posts …

Here’s the data …

Here’s the calcs …

Post # |
Width, b = , in. |
Depth, d = , in. |
Length = , in. |
Volume (cu ft) |
Weight = , lb |
Spec. Wt (pcf) |

1 |
3.51 |
3.51 |
72.5 |
0.517 |
27 |
52.2 |

2 |
3.5 |
3.5 |
72.625 |
0.515 |
26 |
50.5 |

3 |
3.51 |
3.51 |
71.75 |
0.512 |
26 |
50.8 |

Ave = |
51 |

Whoa! … 50, 51 pcf!

The markings on the posts indicate they are `SYP’ (Southern Yellow Pine) No. 2. The `design’ weight for Southern Pine is 35 pcf, but this is for `dry’ service … essentially covered, conditioned structures. For lumber, `dry’ is defined as not exceeding 19 percent moisture content. *(I’ll go check …)* Depending on where I `poked’ the posts, I’m getting high 20’s, high 30’s, 40’s, and in some places pegging out at 50 (percent). There are `equations’ relating moisture content and the specific weight of wood; see, for example, the AITC *Timber Construction Manual* (Chapter 2), or the USDA Wood Handbook (Chapter depending on edition). We could play with the equation(s) provided therein, but what numbers would I plug in, as my moisture content readings are `all over the place’? The point here is: wet wood is heavy! And, if your wood is wet1, and you’re putting it to structural use, or it is supported by some structure, you must consider it’s heavier weight2. Pressure-treated wood is wet because it gets `dunked’, under pressure, to impregnate the, in this case, preservative treatment. It will dry out, some, in time, depending on where I end up using it.

This adventure in wood weight brings me back to what I have `always said’: “Here in America we learn engineering *backwards*.” We learn arithmetic in grade school, algebra and calculus in high school, in college more calculus, and then differential equations, and more numerical methods courses, and at some point start our engineering courses. The courses start with basic mechanics, then analysis, and by the time we graduate, one or several `design’ courses, all the while learning from textbooks.

Then we get a job, and we find out that in the real world, we design using codes and standards and `design manuals’. What about all those (expensive) textbooks? (I’m not suggesting you sell them, or even throw them away … somewhere down the line in your design journey, if one becomes useful for even a few minutes, it paid for itself, if you kept it.) As we trudge forward in our careers, it’s nice if we can see, up close, the materials we’re working with, and the structures we’re designing. Seeing failures is also instructive; hopefully not our own. And maybe we even build a few things ourselves, probably more out of hobby than necessity.

In my journey I have also actually `worked’ on construction crews. Not glamorous, but invaluable. I also worked as an engineer for a construction crew. Double invaluable. And now, even later in career, I am building more and more, primarily structures on our farm. Of late I have become more intimate with the weight of wood, especially the weight of wet wood, dealing with freshly sawn lumber, and pressure-treated lumber. When I was younger, it seemed the wood was always drier; plus, being younger, I was stronger; either way, I `didn’t notice’ as much.

So, in my opinion, an engineering education should be taught in reverse of the way it is now. First, while young and strong, the engineering `student’ should be required to get out there and build stuff … lift, carry, nail. Learn hands-on how much things weigh, what materials are used where, and why, which way things span, how far, and so on. Then enter the `classroom’. Enter the classroom with a real-world situation, with the assignment of coming up with a design, a safe design, a `buildable’ design … a bridge, a building, a … whatever. Be taught the theory, and then `why’ and `what’ captured by the corresponding `equations’. And if fancy math is needed, learn the fancy math. And, finally, the numbers should not be just ink on a page, or computer screen, but reflect reality.

]]>(not the actual house … but close!)

Woeste and Dolan did a paper on floor bounce … wait, no, on floor vibration (copy here). They provided an equation for determining the fundamental frequency of a floor joist, as follows:

f = 1.57 √ (386 EI/WL^3).

Further, their paper says that occupants are the most sensitive to vibrations in the 7 to 10 hz range.

Plugging in the old-house joist dimensions, and span, presumed modulus of elasticity, number of joists involved at any one time, and various weights of movie-set `occupants’ and stuff (for mass) … yeah, it was pretty easy to `land’ in the 7-10 range.

I advised that the floor was indeed safe, despite the bounce … “as long as you don’t put a crazy huge amount of stuff up there … if you want to shoot your movie up there, you’ll have some bounce.”

…

I tried to reconcile the bounce I felt due to the sheer effect of walking across the floor. I `calculated’ that my walking was, say, 1 to 2 hz (steps per second) … not exactly in the 7 to 10 range … but maybe a harmonic of my stepping. Or, each one of my steps engaged more than one joist!

…

To date, that’s been the extent of my movie-set engineering, except for some quick advice on what factor of safety to use for stage rigging.

Over the years I never did see very many 2 x 8 floors. Maybe `bounce’ is why.

I you have a movie for me (engineer, cast, or crew) … let me know!

]]>Consider a conventionally-framed house facing a remodel project. The house is, for the most part, a simple, two-story `box’, with stick-framed roof and interior bearing wall. The interior bearing wall supports both roof and main story floor. Two walls are to be removed per the remodel; one wall is a segment of the interior bearing wall, and the other a non-bearing wall. The interior bearing wall runs longitudinally through the house, roughly down the `middle’. Ten or so feet of the wall are to be removed, opening up the dining room to the kitchen. The segment to be removed, being on the main floor, supports only the roof and ceiling. The non-bearing wall to be removed is between present dining room and living room. Non-bearing walls will be discussed separately. The idea here is to see how we can determine what kind of beam, or header, is needed, to carry the roof and ceiling over this new space, created by removing a part of the bearing wall. A cross-section sketch of the house framing is shown below.

The `easiest’, and perhaps most expensive, way to determine what we need, is to `hire an engineer’. I am such a person. But why? I’m retired (mostly), and since the house was originally designed without an engineer, as are many, why hire an engineer to help with a remodel?!

First, let me say, that you may *have* to hire an engineer. Second, you may *want* to hire an engineer, whether or not you `have’ or `need’ to. Let me explain. A bearing wall is a `structural’ part of the home; it holds part of the home up. Do it wrong and your house may collapse, or look like it’s collapsing. You may *want* to hire an engineer, just for peace of mind. You may *have* to hire an engineer … if the building regulations for your locale require it! This would typically come about via the permit process. You may want, or have to, get a building *permit* to do the remodel. And to be granted the permit, if dictated by the building official, you may have to hire an engineer. In some cases the local building official himself/herself might determine what kind, size, etc. header is needed, or might have design aids for such, but don’t count on this^{1}.

The contractor you choose to do the work (if not yourself) might be willing to determine, or `size’, the header for you. Careful with this one, unless he or she plans to hire an engineer, and pass the associated fees as part of the remodeling contract. While good contractors can do amazing things with buildings and building materials, they typically do not have the science background to properly `engineer’ a solution. And by this I mean a long-term solution … a header that will hold up the roof, not only for the duration of the construction project, but for the `life’ of the structure itself. But, again, if we can come up with a solution without hiring an engineer, let’s do it. And if a permit is required, perhaps your`non-engineered’ solution will make enough sense to the building official, that he / she agrees, and issues the permit. (Yay!) After all, all the headers in the existing building got past having to involve an engineer; why not one or two more!?

Let’s look at some ways to do so.

1. Provisions in the *International Residential Code* (IRC) … online version here. I will be referring to the 2018 edition. Further, the IRC was likely the basis of the building code governing your original structure when first built anyway (if governed at all).

**R602.7 Headers.** For header spans, see Tables R602.7(1), R602.7(2) and R602.7(3).

Table R602.7(1) is for exterior walls … R602.7(2) is for interior bearing walls … and R602.7(3) is for open porches.

The problem with Table R602.7(2) is that it doesn’t seem to accommodate any roof load.

Let’s look at Table R602.7(1) … and … R602.4 …

**R602.4 Interior load-bearing walls.** Interior load-bearing walls shall be constructed, framed and fireblocked as specified for exterior load-bearing walls.

So, let’s use Table R602.7(1) to come up with our header. I suggest that the braces direct the roof load in such a way that the interior bearing wall carries *more* than half of the roof … all of the roof between the braces, and half the spans from braces to exterior walls, or, 5 feet between the braces + ½ of ½ of (30 – 5) on the left and ½ of ½ of (30 – 5) feet on the right, or … 5 + 2 x ½ x 12.5 = 17.5 feet (ft) (of building width). This would correspond to an equivalent (more simple) building width of … 2 x 17.5 ft = 35 ft. We’ll use the first set of `spans’ in Table R602.7(1) … `headers supporting roof and ceiling’. The house is in a region of `no snow’, so, per footnote e to the Table, we’ll use the 30 psf Ground Snow Load. Recall that we want an opening (span) of (about) 10 ft. For a building width of 36 ft we get … 4 – 2 x 12s give us 9-10 (9 feet 10 inches). We should also look at the footnotes, but, before we go any farther … note the `4’. This means FOUR 2 x 12s, each of which is 1.5 inches thick … or a header of 4 x 1.5 = 6 inches. The resulting header will be 6 inches wide (thick) by almost 12 inches deep. That will be hard to hide in an interior bearing wall. Yeah, our building `equivalent’ width is 35 ft, not 36, so maybe we could split some hairs and go a bit smaller (by `interpolating’, per the footnotes), but we won’t be able to reduce it (the header requirement) by much. Even if our building were only 24 feet wide, the requirement (without the weird braces) would be 3 plies of 2 x 12s. Yikes. (That’s why big openings in bearing walls in conventional construction are rare.)

2. Let’s look at provisions in the *International Building Code* (IBC) … online version here.

Chapter 2308 of the IBC addresses conventional light-frame construction … (our stick-framed house) … but we end up in the same dilemma, exactly … 4-ply 2 x 12s. See Table 2308.4.1.1(1). Ugh.

3. Try the *Wood Frame Construction Manual* (WFCM), which is referenced in the IRC, as applicable to (usable for) residential construction (per Section R301.1.1).

Same issues … we don’t see anything directly giving us a solution for an interior bearing wall carrying roof/ceiling only. Tables 3.22A1 and A2 give us options for exterior wall, roof/ceiling only, with the difference in the Tables being dropped header, or raised. A dropped header has a top somewhat below the wall top plates, necessitating cripple studs, which can’t be counted on as providing so-called lateral stability for the top of the header (I’ll explain elsewhere). A raised header is mounted right up under the roof/ceiling construction, stabilizing the top of the header, and is the case of the remodel considered. So we’ll go with A2 … and we see `solutions’ such as 4 – 2 x 12s, roof live load 20 psf, can span 10-8, for building width of 36 feet. We also see 4 – 2 x 10s can span 9-1, and so on … not a whole lot different than what we got with the IRC and IBC.

Big opening … big header required.

Oh, the *Wood Frame Construction Manual* is also available online … here. (I link to the 2018 version, but newer is available, as per the IRC and IBC.)

Let’s look at WFCM Chapter 2. Chapter 2 gives us loads and stresses, instead of specifying the actual header makeup. Look at Table 2.11, upper right. There’s an interior bearing wall holding up roof only! Yay! For a roof live load of 20 psf, building width of 36 feet, the Table gives `720’. This is the unit load on the header or beam or girder. For a building width of 24 feet, we get 480. Going back to the 36 width, then, the 720 means that each foot of header (length), no matter how long, carries 720 pounds. If the header is 10 feet long, the total amount of roof/ceiling load is 720 x 10 or 7,200 pounds (lb). This 720 unit load is also called a `line load’, or pounds per lineal/linear feet, or plf, etc. This doesn’t help us much directly, but can be tremendously useful as we look to some further tools. This number enables us to `engineer’ a header.

Note that if we had a shorter amount of wall removed, we’d be done … try 6 feet in the examples above. Go back and look at Table 3.22A2 in the WFCM … a 2-ply 2 x 10 would carry an opening of 6-3 in a 36-ft wide house. Looking at IBC 2308.4.1.1(1) we see a 2-ply 2 x 10 can carry 5-9, and 2-ply 2×12 … 6-10. Perhaps the difference is that the 2308.4.1.1(1) uses 30 psf snow load for both it and 20 psf live load. And perhaps an eave is considered. Or perhaps there are other effects mixed in (wind?) We’d look closer at the line items in the Manual (or code), and perhaps Commentary(s), if the 6 foot opening is really what we are designing. The IRC comes up with the same thing as the IBC.

4. Design aids provided by industry / manufacturers.

Already you might have realized, justifiably, that dimension lumber isn’t going to work, unless we want a fat header. So let’s look to `engineered’ wood. There are several options. We can go with structural glued-laminated timber (glulam), or structural composite lumber (SCL). And there are others. But, sticking with either glulam or SCL, let’s see what we can find. Let’s see what the glulam people have to offer … the American Institute of Timber Construction (AITC) (on www.aitc-glulam.org). (You’ll get redirected to www.plib.org/aitc as AITC was absorbed by the Pacific Lumber Inspection Bureau (PLIB), and the West Coast Lumber Inspection Bureau (WCLIB) along the way.) Look under Resources, then Publications … and then Southern Pine, and then Roof Beam Construction Load. (Construction load governs where snow loads don’t … consider it the Roof Live Load.) And let’s open up SP 26F – 1.9E, and we have options. Looking at a span of 10 feet … and something that carries (at least) 720 plf, we see … 3-1/8 x 8-1/4 carries 768 B. The `B’ means bending (stress) controls. Whoa, nice … the 3-1/8 width certainly fits in the 2×4 interior bearing wall dimension. And 8 inches is pretty modest. But, we’re not done; we need to be a bit careful. The Table says that the deflection limit is Span/180 for total load. Deflection is a different conversation. But, for now, let’s look back at IRC Section R301.7 and Table R301.7. The Table indicates a deflection limit of L/240 for members supporting flexible finishes, including gypsum, and R301.7 says the limit is based on the … live load of Section R301.6. There is a disconnect … L/180 for `total’ from AITC, and L/240 for `live’ (our construction load) from IRC. We can reconcile the two using Table 1604.3 of the IBC … which relates the L/240 for live load and L/180 for live load plus dead (equaling total).

So, a 3-1/8 x 8-1/4 SP 26F-1.9E (Southern Pine) glulam beam is adequate for the header! (Show that to the building official.)

Now note! … these answers are bare-bones minimums! The AITC Table indicates that the `solution’ is based on bending stress (the `B’). But if we look one column to the right, for the same beam, and an 11-foot opening, the allowable load is 618, and `D’ … that means it’s deflection-controlled. I would argue that the 10-foot span is `almost’ deflection controlled. That would mean that, under the 768 plf, the beam would be expected to have a total deflection of `Span divided by 180’, or 10 feet x 12 inches per foot / 180 = 0.67 inches … about 5/8 of an inch. It might `meet code’, but is that really what you want (to see) above the opening between the kitchen and dining rooms? You might want it to be less, maybe way less! What if we want to drive it down to, say, ¼ inch total. Well, a wood beam is like a spring. We would need a heavier (stronger) spring. Look at the 3-1/8 x 11; it can carry a load of 1365 plf, and is controlled by bending stress. That suggests that, under the 1365, the deflection is not yet at the (Span /180) limit. Maybe significantly under the limit. Since the 3-1/8 x 11 also acts like a spring, under only 720 plf instead of 1365, it would have a total deflection of not greater than 0.67 x 720 / 1365 = 0.35 inches. (Half the load … half the deflection!) Actually less (since bending controls the 1365, not deflection). Maybe way less. (The actual engineering `calc’ not shown.) Turns out to be almost exactly 0.25 inches. Sweet. Generally it’s not a problem going stiffer and stronger than the code minimum(s). Obviously the 11-inch beam will cost more (and weigh more). In a remodel the only issue that we might run into with a deeper beam is `headroom’ … make sure there’s enough clearance under the beam / header to `meet code’ (and your needs … maybe you’re a tall person!).

Let’s try a different product … say Laminated Veneer Lumber (LVL), a type of SCL.

I found that a Roseburg 2-ply 1-3/4 x 9-1/4 2.1E will carry 1307 total (non-snow) load over 10 feet. The 2-ply 1-3/4 is sweet, as the total thickness will be 3.5 inches … perfect for an interior wall (typically 2×4 nominal). Engineered lumber is typically stronger, and stiffer … alas the smaller dimensions. (In the project considered, 2-ply 1-3/4 x 11-1/4 2.1E LVLs were used, and the opening enlarged a bit, to 11 feet). Note that when selecting engineered lumber, you need to make sure it’s `available’ in your locale. Actually, it’s the same with dimension lumber, when dealing with specific species and grade. Southern pine lumber, and engineered lumber made from Southern pine, is generally available on the `South’ (east); Douglas fir, and products made from Douglas fir, are available in the West. And/but there are other lumber species … Hemlock, other firs, cedars, and so on, and some lumbers are `mixed’, e.g., hem-fir, which is a mix of Western Hemlock, and firs. (Just make sure what you want is available, or, find out what’s available, and design within.)

5. Some straightforward calculations and common sense.

Go back to the discussion above where we jump into Chapter 2 of the WFCM, specifically, Table 2.11A. Note the Table heading says Roof Assembly DL (Dead Load) = 20 psf. This is the assumed weight of the roof and ceiling materials (including insulation, whatever). Under RLL (Roof Live Load) of 20 psf, and following the 36-foot width over to the upper right, where it shows a header (or beam) carrying the middle of the roof, we got (get) 720 plf. If we go up to 24-foot building width, we get 480 plf. Now let me show you something. Maybe you already see it. We could say this header (the one in the WFCM graphic) carries half the roof. Or we could say that it carries half the way from the peak to the exterior wall, both directions. Same thing. For a building width of 36 feet, the interior header (or wall) carries an 18-foot wide `swath’ of roof (9 feet each side of the ridge). Each foot of wall (or header) length thus carries 1 ft x 18 ft or 18 square feet of roof. If the roof weighs 20 pounds per square foot (psf), then the wall (or header) carries 18 x 20 = 360 pounds of roof, per foot of wall (or header). Similarly, if the live load on the roof is 20 psf, each foot of wall (or header) carries 360 pounds. Add the two together, and the header (or wall) carries 720 plf, under the design condition of dead weight (or course), and live (probably a re-roofing crew). Yeah, it’s that easy! Similarly, for a building width of 24 ft, the interior bearing wall (and any headers therein) carries ½ of 24 ft x (20 + 20) psf = 480 plf. Determining weights of the material of the roof (dead loads) is not particularly difficult, and the design loads (live, snow, etc.) are typically provided by the local building code … so sizing any header, now, unless the building is unduly complicated, is … doable!

Let’s say we have a building that is 32 feet wide, and the roof is constructed of pre-manufactured trusses (allowed in the IRC, IBC, etc.), that span wall-to-wall. Let’s say we have a design snow load of 60 psf. And we determine that the roof weighs no more than 15 psf. Oh, and let’s add 12-inch eaves. Each exterior wall carries ½ of 32, plus 1 ft, or 17 ft. The total load on the exterior wall (and any headers therein) is 17 x (60 + 15), or 1275 psf. Let’s go back to the AITC Beam Capacity Tables, but this time find one for beams made of Douglas fir, and let’s say we have an opening of 12 feet. Table AITC DF-26 Roof Beams Snow Loads … the 3-1/8 x 12 will carry 1497 plf for 12 feet. Alternately, the 5-1/8 x 12 will carry 1572 plf for 12 feet. Either would work. The 3-1/8 x 15 weighs 11.4 pounds per foot, and the 5-1/8 x 12 weighs 14.9 pounds per foot. Other things equal, the heavier beam will likely cost more. On the other hand, if we’re fighting 60 psf of snow, we’re probably also dealing with 6-inch exterior walls; the 5-1/8 inch wide beam might fit better. Ask the manufacturer; they may well offer widths of 5.5 inches, fitting even better in 2×6 walls. Notice again that the deflection limit is L/180; we could go wider or deeper to get an even `bigger’ (stiffer) beam with less deflection. Notice also that both sizes are controlled by bending stress; thus we know that the deflections will be *less* than the L/180 … how much less we don’t know without some more work. Finally, and importantly, the footnotes indicate that the beams must be supported laterally along the top. This is typically by attachment to a double top plate to which the trusses are attached … holding the top of the beam in place, from buckling out of the plane of the wall. This is the `raised’ header. If the header is lower, a `dropped’ header, with cripple studs from header to the top of the wall, the beam is, strictly speaking, not laterally supported, and there will be lessened bending capacity. A `bigger’ beam may be required. But the bigger beam will give us more deflection control. Some design aids accommodate for the dropped header, either by providing additional tables, such as we saw in the WFCM Chapter 3 Tables, and some aids provide adjustment factors for the tabular values. Otherwise more difficult (`engineering’) calculations may be necessary to deal with the unbraced beam condition. And, of course, make sure this beam is even available in your locale.

So, with this start, you can see some ways to `design’ a header for an interior load-bearing wall. The idea was to avoid hiring an engineer for what are typically non-engineered buildings (residential construction). And we could take what we did here as a start into a plethora of additional ways to design headers … using various online resources, calculators, software, and so on. We could have switched over to steel, and specified a steel beam for the header. Switching to steel might be the only way to deal with high loads and long spans. And, interestingly, though not surprisingly, if you think about it, various manufacturers have design aids by which you can `switch’ from one material to another (typically* their *material) … say a parallel strand lumber (PSL) beam (a type of SCL), to glulam. Or steel to glulam, or … the information out there is endless. No doubt, somewhere, is a table trading dimension lumber for SCL.

At some point, yeah, hire an engineer.

1If it is necessary to involve the building department, make the building official your friend. Who knows what he / she might come up with to assist you on your project.

Aha!, yes, design values for E are average values. Modulus of elasticity is a measure of wood *stiffness.* Wood stiffness is related to how much wood deforms when subject to load, commonly, how much a beam (or joist or rafter) will `sag’. Engineers call this sag `deflection’. We have fancy equations for deflection. The deflection (sag) of the midspan (relative to the ends) of a simply supported beam (or joist, rafter, girder, or similar) is given by the equation:

Δ = 5 W L3 / (384 E I ),

where,

W is the `whole’ load on the beam (joist, rafter, etc.),

L = beam length,

E = the modulus of elasticity of the beam (more to say here),

I = the moment of inertia of the beam, and

5 and 384 are just pure numbers, related to how the load is distributed and the beam supported.

The load W is generally `predicted’, the result of some future event, say a `once-in-a-lifetime’ snow storm. For us engineers this future event is typically predicted by others, and we work off of some amount (e.g., of snow) associated with the event. The number we work with is a prediction, even though it may look like it’s exact.

The beam length, L, and moment of inertia, on the other hand, are more exact. We know pretty darn close how long the beam is (or is to be). The moment of inertia is based on the beam (or rafter, or …) width and depth. We can actually measure these with a tape measure, pretty close, or we can just rely of the dimensions stated (by the manufacturer), plus or minus a bit of tolerance (allowed in manufacturing).

E likewise might very well appear to be exact, such as, E = 1,800,000 psi (pounds per square inch).

Actually, all the zeros betray that, perhaps, only the 1 and the 8 are significant. But, importantly, remember, E is average. Actual E values, board to board, beam to beam, rafter to rafter, etc. … vary. Even for boards, beams rafters of the same species, and grade, and even from the same manufacturer, or even from the same day of* manufacturing.* A common way to describe this variability is through the coefficient of variation, C.O.V. (or COV). It is defined as the standard deviation divided by the average. For a certain beam with a design value for E of 1,800,000 psi, and COV or 0.10, the standard deviation (measure of variability) of the E’s of individual beams of this species and grade is … 0.10 x 1,800,000 psi = 180,000 psi. (As AITC 119 Section 4.1.6 points out) … in a normal distribution, 2/3 of the actual E values `in population’ would be within 1,800,000 plus 180,000 and 1,800,000 minus 180,000 psi. Or, 5/6 of the E’s would be greater than 1,800,000 – 180,00 psi = 1,620,000 psi. Or, half of the E’s would be less than 1,800,000 psi, and 1/6 would be less than 1,620,000 psi. Is that a big deal?

Well, in most cases not. Since E is in the `denominator’ of the deflection equation, a lower E (in an individual, or specific, beam), albeit only a bit lower, will result in a bit greater deflection, or sag (notice I didn’t use the word `higher’). So, if the calculated sag for a beam is ½ inch, there’s a 1/6 chance it could be as much as … 10 percent greater, or 0.505 inches. Sure we could measure the difference; *seeing* it might be more difficult. Only in a case where 0.50 inches (0.500) would be critical, might we worry about this variability. Where specific deflection (sag) amounts are critical, one solution would be to order a stiffer beam (higher E). We could order a beam with E = 2,000,000 psi (if one exists), and calculate the deflection, and its variability. There would still be some `chance’ of the beam having an unacceptably low E, though less chance. If we really want to make sure we have an acceptable E for a particular beam, we need to `measure’ E for that particular beam, though this is typically not done in practice, say, beam by beam, on a jobsite. You* might* be able to get a manufacturer to *measure* the E of individual beams for a particular order … I’ve never done it.

Pictured is a screen grab from the Course, showing an example of conventional roof framing, including the use of a purlin and brace.

The term `conventional’ means `in accordance with general agreement’, `established practice’, and so on. One might also say, with regard to conventional framing … `normal’ construction … `the way we’ve been doing roofs for years’. And this goes two ways in real life: 1) builders who build roofs unencumbered by codes, rules, engineers, etc., … framing them the way they `always have’, and 2) builders who build roofs according to the Building Codes. Those following the Building Codes have two options: one being Conventional, a.k.a. prescriptive, a.k.a. stick framing, and the other, engineered design. Conventional framing is covered by the International Residential Code (IRC), the Wood Frame Construction Manual (WFCM), and Section 2308 of the International Building Code (IBC). Again, the idea being, if one follows the `prescriptions’ in these codes, the employment of an engineer is not necessary. These requirements of these codes seem pretty restrictive, pretty `complicated’, but, I argue, they are generous in that they: 1) relieve you of having to hire an engineer (my contact information is here, somewhere, on the website), and 2) if you follow the requirements, you will not only be following the Code (probably also the law in the area you are building), but you will also be building a `safe’ roof. The requirements of the prescriptive codes are not trivial; nor are they intended to be unnecessarily burdensome or troublesome; they are the product of people coming together to come up with requirements that are both safe and workable in the modern world.

I tell my students, friends, relatives, etc. that are interested in building, or designing buildings, to learn where and how to apply conventional / prescriptive design; so you’ll be able to avoid hiring an engineer. (I am trying to retire.) Conventional design is generally applicable to residential structures. But learn it well, as not all residential structures fall within the limitations of the conventional provisions, and not all `parts’ of some residential structures can be designed (and built) `conventionally’. A bad-case scenario for a designer, or architect, is to embark on a project assumed to be covered nicely by conventional framing requirements, only to find out that an engineer must needs be employed, as the structure, or a part, or parts, fall outside the limitations of use of the prescriptive (`conventional’) provisions.

Another beauty of the conventional framing relates to the term `stick framing’. The roof, however large, can be made up of … `sticks!’ Yeah, pieces of wood that a single person can manage. So, not only may you avoid having to hire an engineer to design a `stick-built’ structure, you can avoid renting a crane, and hiring a crane operator, to put it together. Roof rafters, ties, ceiling joists, and so on, are of manageable size, and also typically available at the local building supply store. Further, the framing pieces are typically attached with nails, as opposed to bolts, or heavy hardware. (Neither do you need a welder!) Where bigger building elements are needed (beams, headers, girders), the prescriptive codes provide for their construction to be of `sticks’, e.g., by nail-laminating them together. (I have found, in fact, that when I have been framing alone, instead of nailing them together on the ground, and having to lift the (heavy) finished beam or header, the header can actually be installed piece by piece. Yay! No crane, no crying for the help of others in the moment!

(So, while I make a living as an engineer designing structures that fall outside of conventional framing, I actually do my living, making structures out of `sticks’.)

(Maybe I’ll pick apart my `conventionally framed’ house in another post.)

Oh, the IRC does provide for the use of roof trusses. The trusses are made up of multiple pieces of lumber, connected, typically, with metal plates (metal-plate connected wood trusses). They typically span from exterior wall to exterior wall, or from intermediate girder trusses to wall, and thus are `bigger’ than single rafters and joists. They are designed by a `truss manufacturer’, and are typically installed with the aid of cranes. You don’t have to hire an engineer to design manufactured roof trusses; the truss manufacturer does. The truss manufacturer typically provides site-specific design and layout, as well as the actual trusses. And the cost of the `engineering’ is part of the `truss package’.

]]>In new construction, wood framing more or less starts from the ground up; wood members are placed on top of one another, and are fastened into place. Just about everything is `statically determinate’, and unless something is cut grossly wrong, things `fit into place’. In remodeling, on the other hand, walls, etc. are removed, surrounding or supporting framing is more or less supported (`shored’), and new construction (beams, etc.) put (sometimes `jammed’) into place. Even with the best intentions, things don’t always `fit’. And when they don’t fit, the common remedy is the hammer, big or small. A common situation is removing a wall placing a header or beam over the new space; the header supporting the framing above, floor or roof/ceiling. While the header (or beam) can be ordered and cut to easily fit, it’s the supports at the ends of the new opening, typically studs (jack studs), that often don’t quite fit.

So let’s look at a recent wall-removed-header-placed `situation’. Roof and ceiling framing were shored using temporary walls on both sides of the existing wall, adjacent to planned opening. The wall framing, from sole plate up to but excluding the double top plate, was removed, and the new header `lifted’ into place (human labor). Some of the jack studs, intended to support the ends of the headers, cut beforehand, didn’t quite fit. Alas the hammer. After a lot of smashing, the jack studs were in place, supporting the header, readying for the removal of the temporary walls. Not quite!

Let’s run some numbers.

The header was designed to have very little `sag’ … `deflection’ in engineer language (as well, of course, be strong enough). The existing wall held up ceiling and roof framing above; the new header would take over the role of the wall above the new opening. The clear span for the header is 10’-9”. Two jack studs were used for support each end. In the design headers, beams, etc., the `structural’ span for the header is taken to extend half way into each support, so, 10’-9” + 2 x ½ of 2 x 1.5” = 10’-9” + 3” = 11.0 ft. The header is a 2-ply 1.75-in. x 11.875-in. 2.1E LVL. The `design’ of the header accommodated the design load by several hundred percent, with, importantly, very little anticipated deflection, or `sag’ (very stiff header). The intent was that the top of the opening be visually `flat’.

The jack stud rammed into place at one end caused the header to be `bent’, with respect to a straight line end to end, about 1/16th of an inch. (Lots of ramming). This `bent-ness’ was before the shoring was removed. Properly shored, and assuming a flat (straight) ceiling/roof above, the header initially should `fit’ under the framing with no deflection; it’s not carrying any load! Only as shoring is removed should the header take on load, and begin to bend. Structurally, the action of forcing the header to be bent into a curved shape, however `slight’, produces, in the wood, the same effect as if the header was simply supported at ends (just `resting’ on supports), and being loaded downward along the span, enough to bend it. The load comes about as jamming the jack stud(s) in place tends to `lift’ the header and framing on top of it … the framing above resisting the lifting action, pushing down.

Assuming this `lifting’ effect is greatest at the end where the stud is jammed in place, and tapers off to … nothing, or little, at the other end, we’ll model the situation using the so-called `SIMPLE BEAM-LOAD INCREASING UNIFORMLY TO ONE END’, found in typical engineering design manuals. (See graphic.) Let’s look at the load `required’ to produce a 0.10-inch deflection.

Δ = 0.01304 W L^3 / EI,

where,

Δ = deflection (sag, out-of-straightness),

W = the `whole’ load (`weight’) or downward force on the beam (header),

L = header span, 11 ft or 132 in.

E = the modulus of elasticity of the header, 2,100,000 psi (the `2.1E’), and

I = area header moment of inertia of the header , n x (b h^3/12) = 1.75 (11.875)^3 / 12,

where

n = number of plies,

b = thickness of each ply, and

d = depth of the header.

We get, for I, 488 in.4.

Solving for W, we get, … W = 2,137 lb (call it 2,100 lb).

Hmmmm. The total dead load (roof and ceiling weight) above the header is about 2,300 lb, and more or less distributed uniformly. By smashing the jack stud in place we are just about starting to `lift’ the roof. No. Take the jack stud(s) out, take off a sixteenth of an inch or so, so that it goes in with less `smashing’.

*Can’t we just leave the header as is, since it, and the jack studs, are in place, and nothing `happened’? After all, the header was intended to carry the roof weight anyway!*

Maybe, but why? What if the out-of-straightness is 1/10th of an inch? … or 1/8th of an inch? At 1/8th of an inch we’d be well overcoming the weight of the roof … things would start to `give’ that we don’t want to give. Additional to the noise from the banging of jack studs in place, we might be able to hear creaking and groaning as the framing (wood members and fasteners and other materials) above and adjacent to the header start crushing, yielding, cracking. At some point we could, theoretically, `break’ the header itself (if we haven’t already broken something else … maybe the jack stud). Besides, we’d be carrying the roof wrong … by the end of the header, not equally along the entire span.

Are you SURE nothing happened? Is the roof plane still straight? Have you cracked any gypsum?

The header is now pre-bent, and pre-loaded. Is the out-of-straightness visible? So far the header is only carrying the weight of the roof itself, plus some more. What happens when the header experiences its `design load’ (a future re-roofing project)? In this particular case, re-running the design calculations with some extra load (and deflection) might turn out to be `okay’ (as there was way more strength, and stiffness in the header than required), but, again, why? Take the studs out and do it right. The header should rest snug, tight, on top of the studs, but not so jammed as to deflect the beam.

…

Where a header or beam is more deeply embedded in a structure, and, particularly, if the structure has continuous framing elements or inflexible systems, providing little `give’, it’s possible, by `jamming’ studs into place, to induce a load in the header that exceeds the design load. Not good.

Don’t ruthlessly smash studs or columns into place; cut them so they fit. Tight, yes! And with a bit of pounding; but you don’t want the header ends to `settle’ (move) when the shoring is removed.

…

A NOTE ON NUMBERS

The numbers in the calculations above look rather `exact’. Actually, many are not. For example, the modulus of elasticity value, E, for the header is (published) 2.1E, meaning 2,100,000 psi. This is an average value. A bunch of LVL specimens were measured in a test lab for E, and the average value was used (published). Some of the LVLs that arrive on a project (or lumber yard) have higher E, some lesser. That’s the way wood is, variations in properties, between species, between grades in species, and between `engineered lumber’ (e.g., LVL) made up of wood. (LVL tends to have less `variation’ than other wood products, due to how it is manufactured … that is another conversation.) So, when I dump and un-exact number E into an equation that looks exact (and, in fact, might very well be), I get an un-exact number out. The applied force to bend an 11-foot long 2-ply 1.75 x 11.875 2.1E LVL header 1/16th of an inch may take somewhat more, or somewhat less, than the one I calculated; in fact, it would be highly unlikely that it would take, exactly, 2,137 lb to bend the header 1/16th inch, exactly.

Further, the deflection equation used does not account for `shear deflection’. Shear deflection is the relatively small amount of additional `sag’ that a beam, header, joist, girder, whatever, experiences due to the `shearing’ of the wood fibers. The equation used above only determines the deflection (sag) that comes about from bending stresses (compressing the top of the beam and elongating the bottom). By neglecting the shear-deflection term I have been a bit `conservative’ – a lesser amount of force (jamming) could indeed produce the same effect. (The header is a tiny bit less stiff than I calculated.) But, this `shear deflection’ may very well be (more than) eclipsed by the variation in E, already discussed above (and talked about more in some of my other posts).

Don’t worry so much about the exactness in E; worry about cutting the studs so that they don’t need to be ruthlessly smashed into place.

…

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