# what are the chances?

In a previous post (here), we determined the size glulam roof beam required to *not* deflect more than 1.00 inches. We arrived first at a size 6.75 in. x 27 in., but it was not quite sufficient. So we ended up going to the next size bigger, 6.75 x 28.5 in. The design load on the beam was from snow (and a lot of it), and dead (roof/ceiling material weights), plus the weight of the beam itself. Assuming the contractor builds the roof according to the architectural and structural plans, which detail the materials involved, our `dead load’ numbers should be pretty accurate. But the snow load we had to … `predict’. We want the roof to withstand some future snow event, which (since it hasn’t happened yet), is, strictly speaking, unknown. So, for future snow events, we rely on the expertise of others who know a lot about snow, and numbers, and they provide `amounts of snow’ and probabilities of occurrence. In this example, we are using a `50-year’ snow event. Loosely speaking, it’s the amount of snow that occurs `once in 50 years’. Strictly speaking, it’s the amount of snow that has a one-in-fifty chance of occurring in any given year (most often the `year at hand’). And, strictly speaking, it’s not just the exact amount of snow, but the amount *or greater*. In other words, if the 50-year snow is 67 pounds per square foot (psf), then there’s a 1-in-50 chance, each year, that this amount will be equaled or exceeded. (Or, we can expect a snowfall of 67 psf, or greater, to occur `once in fifty years’.) Enough said.

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 …

*11 percent*more than needed. (Hold this thought!)

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’.