# Absolute Deflection

Let’s look at a situation where, after construction, it is required that a certain roof beam experience a deflection of not more than 1.0 inches (in.). The deflection limitation comes about by a required clearance between the ceiling and some `furnishing’ installed in the space below (bank of computers, shelving, whatever). The ceiling is supported by a system of beams, each spanning 32 feet (ft), and spaced 12 feet apart. The beams support (carry) roof and ceiling construction, as well as other loads imposed on the roof, in this case snow and wind.

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 L4 / 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 L3 / 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 L3 / 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 L3 / 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) loads2. 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,

fb = M / S where

S is the section modulus, b d2/6 = 6.75 in. (28.5 in.)2 / 6 = 913.8 in.3.

Thus, fb = M / S = 1,528,320 lb-in. / 913.8 in.3 = 1,672.5 psi.

The allowable bending stress is

Fbʹ = Fb 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, CV.

CV = (5.125 / 6.75)1/10 (12 / 28.5)1/10 (21 / 32)1/10 = 0.855.

Thus, Fbʹ = Fb CV = 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.