If we go back and look at the calculations, we needed a beam moment of inertia (I) of 11,770 in.^{4}. The 6-3/4 x 27 provided 11,072 … barely * not* enough. The next size bigger is/was 6-3/4 x 28-1/2, which has a moment of inertia of 13,031 in.4 … more than enough …

Now let’s go back to the original calcs. Deflection is based on the load (snow plus dead), the size beam, the *kind* of beam, how the load is distributed along the beam, and how the beam is supported (simple span, cantilever, multi-span, etc.). In our case we have a simply-supported beam under uniform (uniformly distributed) load. The deflection is/was calculated using the equation

Δ = (5 W L^{3}) / (384 E I) …

where W = the whole’ load, in our particular case snow (S), plus half the dead load (0.5D)^{1},

E is the modulus of elasticity of the beam,

I is the beam moment of inertia (function of its `size’), and

the 5 and 384 reflect the load distribution (uniform) and beam supports (at ends, `simple’).

We can `dial in’ on nearly all of the `ingredients’ to the equation above, in the situation described, except for S and E. The amount of snow is a `prediction’. And for E we generally use a published *average* value. Beams coming off the production line have actual E values of less, and more, than the published average. By using a published E value of 1,800,000 psi, we have a 50 percent chance of getting a beam with less E (more deflection), and 50 percent chance of getting a beam with more E (less deflection).

In any given year, then, what is the chance we’re going to get a deflection of 1.00 inch, or more? …

Two percent (snow) times 50 percent (stiffness) … ONE PERCENT CHANCE. Albeit we might get pretty close!!!

What is the chance we’ll get the 50-year snow at least once … *over a 50-year period?*

One in 50? … NO!

As detailed in other posts (albeit stream and flood flows) … we need to look at non-occurrence.

What is the probability that we have NO 50-year snow events for the next 50 years?

The probability of the 50-year snow *not* occurring in any one year is 98%, or 0.98.

The probability that we will not experience^{2} that event in the next 50 `consecutive’ years is … (0.98)^{50}, or 0.36, or 36%.

There’s only a 36 percent chance we’ll *not* see the 50-year event during the next 50 years^{3}.

Thus, there’s a 64% chance … that we WILL! It’s `more probable than not’ that we’ll experience at least one `50 year’ snow, … in a 50-year span.^{4} Almost 2-to-1 odds that we *will* experience the event, at least once in 50 years.^{5}

So, considering the next 50 years, and the variability of E, there’s a 0.64 x 0.5 or 32 percent chance we’ll hit 1.00 in. deflection, or more, if our beam has exactly 11,770 in.^{4} of moment of inertia, with, strictly speaking, unknown E (for which we used an `average’ value). But … our beam doesn’t have the perfectly needed 11,770 … it has either 11,072 (smaller, 27-inch-deep beam, for which I argue we’ll have a greater chance of hitting 1.00), or 13,031 (28-1/2-inch-deep beam), which I’ll argue has *less* chance of hitting 1.00 inch.

Here goes …

First let’s combine E and I into a single `stiffness’ term.^{6} To get `exactly’ 1.00 inches of deflection, we need an `I’ value of 11,770 in.^{4}, assuming E = 1,800,000 psi … or EI = 11,770 in.^{4} x 1,800,000 psi = 21,186,000,000 lb-in^{2}. If we need 11,770 in.^{4}, but have 13,031 in.^{4}, for the same EI, we only need … 21,186,000,000 / 13,031 = 1,625,815 psi … for E.

Q: What is the probability that our beam has at least 1,626,000 psi?

A: Assuming that the modulus of elasticity, E, of a certain grade beam, coming off the production lines, follows a normal distribution …

Z = (x – μ) / σ, where

Z is a function in the normal distribution (`Z statistic’),

x = is the value under investigation, 1,626,000 psi,

μ = is the population mean, in this case 1,800,000 psi, and

σ = is the standard deviation of the population (of E values).

For structural glued laminated timber, mean properties are often described in terms `coefficient of variation’ (COV), which is the standard deviation divided by the mean. For glulam, the COV for modulus of elasticity is taken to be 0.10, so,

σ = 0.10 x 1,800,000 psi = 180,000 psi.

Then Z is … (1,626,000 – 1,800,000) / 180,000 = – 0.968.^{7}

The corresponding probability of having E values less than 1,626,000 psi (Z = – 9.68), from a `Z Table’, is 0.167 … 16.7 percent.

So, there’s a 50-50 chance of getting a beam with less than 1,800,000 psi, and a 17/83 chance of getting one with 1,626,000 psi.

But, since we have more than enough I, then an E of 1,626,000 psi still works!

The probability, over the next 50 years, that we’ll experience the `big snow event’ (67 psf), is 64 percent (more probable than not), but because we have extra I, we can afford less E … in fact, it’s more probable than not that we’ll have *extra* E, so much so, that the probability of experiencing 1.00 inch (or more) deflection is …

0.64 x 0.17 = 0.11 … 11 percent … (way) more probable than not that we will *not* have a deflection problem. There’s only an 11 percent chance we’ll hit (or exceed) 1.00 inches of deflection (over the next 50 years).

NOTE 1: We looked at the `chances’ with respect to *one* beam, and only with regard to deflection. We cannot say the same (11 percent chance) for other properties or design conditions, and certainly not for the whole building.

NOTE 2: Design properties of wood products such as sawn lumber typically have higher COV values than those for glulam.

ENDNOTE: If it’s really important that the 1.00 inches is never exceeded, even by a little, you must consider the variability of E, and the probability of `S’.

1In examining sufficiency of the *strength* of the beam, we most certainly consider ALL of the dead load. In considering the deflection due to the ‘S + 0.5D’ we’re assuming that the roof-ceiling-beam assembly has deflected due to its own weight, and then something gets built under it requiring 1.00-inch clearance. As time goes on, the roof-ceiling-beam assembly deflects a bit more due to its weight (0.5D), and some time after that, we get the big snow (S).

2The years don’t even have to be consecutive, but that’s how we march through life … this year, then the next, and so on.

3`Climate change’ notwithstanding???

4When I first started out as an engineer … a `50-year’ snow was some future, almost `mystical’ amount, and nearly never, event, that I could only imagine. Now that I have been doing engineering for 50 years … I’ve seen *several* `50 year’ events … even a few `100 year’ ones.

5NOT 100 percent (certain) …

6E and I are often combined into a single stiffness term, EI, for engineering wood products, such as prefabricated wood I-joists.

7This shows us that 1,626,000 psi is about `one standard deviation less than the mean’.

In the determination of wood beam stability (lateral torsional buckling) with respect to bending about the X-X axis, use Ey min (not Ex min).

The context here is structural glued laminated timber (glulam). And it’s a bit confusing, since Ex min is under the heading `loads perpendicular to the wide faces of the laminations’. But, I don’t know of a situation where Ex min would be used when the beam is loaded perpendicular to the wide faces of the lams (X-X bending).^{1} The beam `buckles’ the other direction … mandating the use of Ey min.

1One might think … `ponding’. *Maybe* … but remember, Ex min has the shear component of deflection stripped out, and ponding is definitely a ponding situation. The AITC *Timber Construction Manual *utilizes E05 (with shear component retained).

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 (strength), shear (strength), 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’ drives the design values (for bending, shear, etc.) 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 (or 1.05 for glulam) 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 = 1,900,000 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 an `actual’ E 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 the `Z statistic’.

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 below 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 E of (less than) 950,000 psi.

Almost `never’.

I have often wondered what this beam, or board, would `look like’? (Would it even make it out of the manufacturing plant? I doubt it.)

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 (piece) 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 huge amount of wane? Some other series defect(s)? Would it even make it onto the grading table? Maybe? Would it make it off the other end?

So, we’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 wood product, like glulam … because of the COVE. While we would essentially `never’ run into a glulam beam with and E actual of Emin, we *might very well … *with sawn lumber.

First let’s see if `normal’ deflection criteria used in beam design will be sufficient … by this I mean the criteria that we typically use in engineering design / structural design software. The criteria is found in IBC (International Building Code) Table 1604.3. For roof members `supporting plaster or stucco’, the deflection due to live (in this case snow) load is limited to *l*/360, *and*, the deflection due to live load plus the creep component of dead load … (is limited to)* l*/240 (where *l* is the member span). The ceiling finish material is gypsum board, which I will assume is crack-able … `plaster or stucco’. Thus, before we even know what the actual design loads are, the `code’ deflection limits are:

*l*/360 = 32 ft x 12 in. / ft divided by 360 = 1.067 inches, and

*l*/240 = 32 ft x 12 in./ft divided by 240 = 1.60 inches.

So, in this case, the 1.0 in. limitation is more stringent than the `code minimums’. (And that’s fine; we can always be more *stringent* than the `code’, but not less.)

Next we look at the actual loads on the beam, and determine a size `big enough’ to hold the deflection (sag), to 1.0 inch (or less). In this situation the design loads for the roof are 67 pounds per square foot (psf) snow load, and 12 psf roof dead load (excluding beam weight).

The loads arrive on the beam as `line loads’, and are calculated to be …

… 67 psf x 12 ft (tributary width) = 804 pounds per linear foot (plf) (of beam) snow load, and

… 12 psf x 12 ft = 144 plf dead load.

First let’s look at the effect of the snow load. In this example the line load is uniform.^{1}

The (`single-term’) equation for deflection of a (simply supported) beam subject to uniform load, in terms of uniform line load, is

Δ = (5/384) w L^{4} / EI,

where

5/384 is the theoretical constant based on load distribution and support conditions (in this case a simple beam under uniform load)

w is the uniform line load, 804 plf snow, in this first calculation,

L is the beam span,

E is the beam modulus of elasticity (measure of beam material stiffness), and

I is the beam moment of inertia (a measure of how `big’ it is).

We will investigate a suitable structural glued laminated timber (glulam) beam, for which E and I are addressed separately. For some wood products, E and I are combined to a single `stiffness’ term, EI.

I prefer to use the above equation cast a bit differently, where w is multiplied by L to get the `whole’, W, load on the beam … w L = W …

Δ = (5/384) W L^{3} / EI.

Since in this investigation we have specific deflection criteria that we are looking at, but, so far, unknown beam (size), let’s cast the equation in terms of … `needed (beam) I’.

I (needed) = (5/384) W L^{3} / E Δ.

The `whole’ snow load on the beam is 804 plf x 32 ft = 25,728 lb.

L = 32 ft or 384 inches.

A commonly specified glulam beam is the 24F-1.8E … in the West it will typically be manufactured from Douglas fir / Larch; in the East, of Southern Pine. The 24 represents a bending stress design value of 2400 pounds per square inch (psi), which we will talk more about later, and the 1.8 represents a modulus of elasticity of 1,800,000 psi.

To limit the snow load deflection to 1.00 in., we will need a moment of inertia, I, of …

I (needed) = (5/384) W L^{3} / E Δ = (5/384) 25,728 lb (384 in.)^{3} / (1,800,000 psi x 1.00 in.) =

I (needed) = 10,538 in.^{4}.

That’s a lot of moment of inertia. Yeah, but that’s a lot of snow … 12 tons! … that’s a heavy snow load … tons and tons of snow … probably 6 or more feet of snow on the roof above.

Looking at `Western’ (Douglas fir / Larch) glulam, some available sizes are … along with moment of inertia values …

6-3/4 x 25.5 … I = 9,327 in.^{4},

6-3/4 x 27 … I = 11,072 in.^{4}, and

6-3/4 x 28.5 … I = 13,031 in.^{4}.

The 27-inch-deep beam is deep (big) enough.

So, once the building is constructed, the beam can accommodate 67 psf of snow (assuming strong enough), *and* accommodate the 1.00 inch deflection limit. (In fact, it can *more* than accommodate the load, as it has 11,072 in.^{4}, versus the 10,538 needed (5% extra).

But wait! … there are a couple things we need to look at.

First, the 1,800,000 psi modulus of elasticity for the 24F-1.8E is an *average* value. Some beams will come off the production line with greater E, and some with lesser. Greater E beams we are generally not going to worry much about, but if we have an absolute limit of 1.00 inches, and `half’ the 24F-1.8E glulam beams actually have *less* than 1,800,000 psi, we need to look at just how serious we are about the 1.00-inch limit. What is the probably we’ll have a beam with lesser E? (That’s easy … 50/50.) And, importantly, what’s the probability we’ll actually experience a snow load of 67 psf? We’ll talk about probabilities later.

*Second*, glulam beams `creep’ under sustained (long-term) loads^{2}. They deflect (sag) initially, and as time goes on, they sag some more. For glulam, this additional sag is taken to be 50% of the initial, and typically, the `sustained’ loads involved are the dead loads (weights of roof construction). Any additional sustained loads must also be considered, such as HVAC, and other `long-term’ appurtenances. (In this example we’ll assume none.) The design snow load is assumed to be transient (temporary), and does not enter into the creep determination.^{3}

For the 6-3/4 x 27 `Western’ glulam, I find a beam weight of 44 plf.

The uniform dead load on the beam, including the beam weight, is 144 + 44 = 188 plf.

The `Whole’ dead load plus beam weight is 188 plf x 32 ft = 6016 lb.

There are several ways to `handle’ this.

Perhaps the way most `reflective’ of the code would be to find the deflection of the beam due to snow plus `half’ of the dead load …

W = S + 0.5D = 25,728 lb + ½ of 6016 lb = 28,736 lb.

Now let’s find the I needed to limit this snow plus creep effect to 1.00 inch …

I (needed) = (5/384) 28,736 lb (384 in.)^{3} / (1,800,000 psi x 1.00 in.) = 11,770 in.^{4}.

Yikes! … this pushes us into the next bigger beam size (6-3/4 x 28-1/2). `One more lam’. If we want to limit deflection to 1.00* for the long term*.

…..

Before we talk about probabilities … let’s first make sure the 6-3/4 x 27 … oh, in this case 28.8 … is *strong *enough.

Bending stress … the bending stress in the beam under the design load will be snow plus the *full *dead load, and weight of beam.

The 28.8-inch deep beam is a bit heavier … weighing 47 plf (I looked it up).

Total uniform load, w = 804 plf + 144 plf + 47 plf = 995 plf.

W = 995 plf x 32 ft = 31,840 lb.

For a uniformly loaded, simply supported beam, the maximum bending moment is,

M = W L / 8 = 31,840 lb (32 ft) / 8 = 127,360 lb-ft = 1,528,320 lb-in.

The extreme fiber bending stress is,

f_{b} = M / S where

S is the section modulus, b d^{2}/6 = 6.75 in. (28.5 in.)^{2} / 6 = 913.8 in.^{3}.

Thus, f_{b} = M / S = 1,528,320 lb-in. / 913.8 in.^{3 } = 1,672.5 psi.

The *allowable* bending stress is

F_{b}ʹ = F_{b} multiplied by the appropriate adjustment factors.

Fb = is the bending stress design value, 2400 psi.

The beam is fully braced laterally, so the only adjustment factor in this case is the Volume factor, C_{V}.

C_{V} = (5.125 / 6.75)^{1/10} (12 / 28.5)^{1/10} (21 / 32)^{1/10} = 0.855.

Thus, F_{b}ʹ = F_{b} C_{V} = 2400 psi (0.855) = 2053 psi.

Since the bending stress under full design load of 1672.5 psi is less than the allowable bending stress 2053 psi, the beam is strong enough. (I’ll leave the shear stress check up to you.)

Okay. Remember, there’s a 50/50 chance that the actual modulus of elasticity is less than the 1,800,000 psi used in our calcs. A couple questions remain. What is the probability of actually experiencing the 67 psf snow load used in these calcs? And, what kind of cushion does the `extra’ moment of inertia, I, provide (since we needed 11,770 in.^{4}, but had to jump up to a beam with 13,031 in.^{4})?

We’ll address these questions next!

1In many cases snow loads are *not* uniform, due to drifting or other effects.

2As do wood beams in general.

3Snow accumulations may persist on roofs for significant amounts of time, but not necessarily at the maximum (design) load amount.

Next comes `dimension’ versus `dimensional’. As an engineer I deal with `dimension’ lumber. Dimension lumber is the term used in the model building codes, and the *National Design Specification® for Wood Construction* (NDS). You won’t find the word `dimensional’ in the IBC (*International Building Code*) or IRC (*International Residential Code*) … at least not in reference to lumber. Conversely, you might not be able to find `dimension’ lumber in the building supply store, even though they may have plenty of dimension*al *lumber. As an engineer, I used to let it bother me when someone used the term dimension*al* to describe dimension, but I’ve eased up a little. Even some engineers, and an educator or two, use the wrong term.

And then there are stud versus studs. To a builder, a `stud’ is a vertical framing member in a wall. One might say `any’ vertical framing member. Common sizes of studs are 2 x 4 and 2 x 6. But there is also `stud’ grade of lumber. We could even say stud grade of stud. As an engineer designing a wall to meet certain strength requirements, I often specify a grade of `No. 2’ for the studs. It’s often available, and has reasonable strength and stiffness. On the same aisle with the No. 2 grade lumber (studs), there might very well also be `stud grade’ studs. These stud grade studs have significantly less strength and stiffness. (Yikes.) So, if a set of plans says `No. 2’, use No. 2.

Then there’s `select’. When an engineer needs all the strength available in a size and species of lumber, he/she might specify `select structural’. Among other grades of the same species and size, select structural will be the strongest. It will also likely be at the top price-wise, and may or may not be available. But, be careful, `select’ may mean other things in the building-supply store. Select may refer to *appearance*, not strength. Or it may be part of a brand name.

Dimension, common, select … those stand out as the most confuse-worthy … the `biggies’. There are others. I used to get all wound up about the mix-ups, but now just smile when the mix-up is someone else’s, and just try to be as clear and deliberate as possible with anything I am designing.

Fortunately, in the IRC, stud grade lumber is allowed for vertical framing members, a.k.a. `studs’ (Section R602.2) (unless specified otherwise).

]]>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

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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!

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