# Deflections of Wood Beams

When we examine deflections of wood members (beams, joists, rafters, girders, etc.) we generally look at two kinds, actually three … the deflection due to the `live load’, e.g., people walking on a floor, wind on a wall, and so on, … the deflection due to the `dead load’, e.g., weights of the construction materials, i.e., weight of the floor itself (dead weight) … and … the long-term additional deflection due to the sustained loads (dead or live, but generally dead). There is the immediate effect of a load, and the long-term effect. Generally the long-term effects are associated with the dead loads, as the live loads are more `transient’, i.e., short-term, come and go. The dead loads are generally the weights of the building materials themselves; these loads don’t `come and go’, as do loads from occupants, wind, snow, etc. When wood members (beams) are subject to short-term loads, they deflect, and when the load goes away, the beam un- deflects … essentially goes back to the un-bent or pre-deflected shape. But if the load stays on the beam for a while, the beam deflects, and then deflects more. This is called `creep’.

And, further, when off-loaded, the beam does not go back to its un-deflected shape.

Example

Let’s look at an example. Consider 2 x 10 floor joists, spaced 16 inches on center, spanning 12 feet (ft) `simply’, to be used in the main floor of a residential structure. The International Building Code (IBC) requires that the joists be able to carry (be designed for) an occupancy load of 40 pounds per square feet (psf). The joists, ultimately, will need to carry: 1) this occupancy load (people, whatever, for the most part coming and going); 2) the weight of the floor system; and 3) the weight of the joists themselves. In the case of joists, their weight is often included in the weight of the floor system (i.e., floor made up of joists, sheathing, finish floor materials, gypsum ceiling, perhaps thermal or acoustic insulation). On the other hand, the weights of beams, girders, etc. carrying joists, should be specifically considered. The dead weight (load) of the a typical wood-framed floor system seldom weighs more than 10 to 15 psf. If the floor construction includes tile, or concrete, then, obviously … more.

Live Load Deflection

The live (occupancy) load that each joist is required to support can be calculated two ways. I like to look at the `whole’ live load. The `whole’ load is the `area load’, in this case 40 psf, times the `whole area’ (floor area) carried by the joist (not the whole floor, just each joist). Floor joists are spaced 16 inches (16/12 or 1.33 ft); so, each joist will need to carry (support) … 1.33 ft x 12 ft = 16 square feet (sf). The `whole’ live load is, thus, 40 psf x 16 sf or 640 lb. You could think of this as several people ending up standing on the joist at one time.

The general formula for calculating deflection of a simple beam supporting a uniform load, in terms of `whole load’ is …

Δ = 5 W L3 / (384 E I),

where

Δ = the calculated deflection,

W = whole load,

L = rafter span,

E = the moment of inertia of the wood joist, and

I = the moment of inertia of the wood section, and

5 and 384 are `numbers’ particular to the beam loading and support condition, in this case uniformly distributed load on simply supported beam (joist).

Another common form of this equation is …

Δ = 5 w L4 / (384 E I),

where

w (`lower case’) = the load in terms of load per length of joist.

It’s pretty easy to see that W = w L.

This `line load’ w can (also) be calculated as,

w = σ x s,

where

σ = area load (in psf),

s = tributary width for the joist (joist spacing), and, of course,

x is the multiplication symbol, in this case.

(Let’s check: w = 16/12 ft x 40 psf = 53.3 lb/ft (plf); W = w L = 53.3 lb/ft x 12 ft = 640 lb. Yay!)

Continuing,

(So, W = 640 lb.)

Let’s get L into inches … 12 ft x 12 in./ft = 144 in.

For E we need to pick a wood species, and `grade’ within the species. This will depend on availability/cost. Availability depends on location. Around here (Southeastern USA) Southern Pine is typically available. A good starting place for cost is `Grade No. 2’. A good place to get the E for wood products used in construction is the Supplement to the American Wood Council National Design Specification® for Wood Construction (NDS). You can view it online at awc.org. It’s a good place (good source) because the NDS is referenced by model building codes, and thus likely acceptable to the building department with jurisdiction over `your’ project. (Plus you can view it `free’.) From Table 4B in the 2018 NDS Supplement we get … E = 1,400,000 psi.

For I we use the formula for the area moment of inertia for a rectangular section,

I = b d3 /12.

Assuming the joists will be installed `on edge’ and from Table 1A in the (same) Supplement, b = 1.5 in., d = 9.25 in., and, thus,

I = 1.5 in. (9.25 in.)3 / 12 = 98.9 in.4.

(We could have just looked I up in Table 1B of the `Supp’.)

Here goes, … the calculated deflection for the 40 psf live load is …

Δ = 5 W L3 / (384 E I) = 5 (640 lb) (144 in.)3 / [ 384 (1,400,000 psi) (98.9 in.4) ] = 0.180 in.

The `bare bones’ minimum deflection required for live load deflection of a floor joist, per Section 1604.3 of the IBC, is that the deflection due to live load shall not exceed the span divided by 360. By `bare bones’ I mean that this is a minimum requirement. The owner, designer, floor material manufacturer, etc. may require or recommend something more stringent … say L/480, or L/600, or an actual (fraction of an inch) amount. The manufacturer of a brittle floor material may, for example marble, recommend a more stringent deflection. (Not to mention it might very well weigh more, motivating a more deliberate calculation of the floor material weights.) But don’t try going any less stringent, or an occupant might be able to feel the beam deflect; or an observer see the floor deflect; or more deflection might cause a problem with floor or ceiling coverings or materials. In the case of our example,

L / 360 = 144 in. / 360 = 0.40 in.; 0.18 in. ≤ 0.40 in.; by this criteria we are okay.

Dead Load Deflection

The IBC places a limit of L/240 on the additional deflection due to dead load, plus that due to live load. And the additional deflection due to dead load may be taken as the immediate (direct) deflection of 0.50 times the dead load (0.5D) (for dry lumber in dry condition of use). It’s another way of saying that the additional deflection due to the long-term (dead) load is 1.5 times the immediate, or 50% more.

Let’s do it.

Our `whole’ dead load is …

W = 15 psf x 16 sf = 240 lb.

Δ DL = 5 (240 lb) (144 in.)3 / [ 384 (1,400,000 psi) (98.9 in.4) ] = 0.067 in. (CHECK THIS!)

The deflection of `0.5D’ would be half that, or … 0.033 in.

Added to the live load deflection of 0.180 in., gives 0.213 in.

Comparing this with L/240 = 144 in. / 240 = 0.60 inches … we’re good!

The TOTAL dead load deflection, long term, is 0.066 in. + 0.033 in. = 0.099 in., which `persists’. Then the occupants arrive, and we get 0.180 inches more, or 0.099 + 0.180 or 0.279 total inches. The occupants leave, and the deflection `goes back’ to 0.099 … (assuming the occupants don’t stick around `long term’).