# E (modulus of elasticity) of Wood

The modulus of elasticity (MOE, E) of wood is most commonly used in the calculation of deflections (sag) of wood beams. By `beams’ I include joists, rafters, girders, decking boards, any … wood used structurally to support loads across spans or spaces between other beams, foundation walls, posts, or other points of support. A commonly used equation is that for determining the deflection of a `simple’ beam supporting a uniformly distributed load between its ends. The equation can easily be found in engineering textbooks, structural design aids, and so on. Here is one form of the equation:

Δ = 5 w L4 / (384 E I),

where

w = the uniformly applied load, i.e., such as in pounds (lb) per foot (ft) (of beam length), or plf,

L = beam length, i.e., in ft, or inches (in.),

E = the modulus of elasticity of the wood, typically psi (psi per inch of deflection per inch of wood),

I = the area moment of inertia of the wood section, i.e., in in.4., and

5 and 384 are just numbers (pure, exact numbers), based on the support and load distribution on the beam.

We call this `case’, or condition, the simple beam – uniform load, or `uniformly loaded simple beam’.

The equation is dimensionally consistent; it can be used in any `system of units’, as long as done so correctly (e.g., get all the input into inches and pounds).

I tend to like a slightly different form of the equation, commonly used in the structural steel industry, but it still works for wood. Here it is …

Δ = 5 W L3 / (384 E I),

where

W = the `whole’ load on the beam.

(The `whole’ load on the beam is simply the load per foot (or whatever unit) of length, times the whole length … W = w x L.)

Let’s do an example. Let’s say we intend to use a 2 x 10 dimension lumber joist to support a floor. The floor is estimated to weight 15 pounds per square foot (psf), including joists, and is intended to carry a design occupancy load of 40 psf. The joists will be spaced 16 inches apart.

The joist must carry the floor itself, plus the occupancy load (people, furniture, etc.); thus the total load applied to the joist will be 40 psf + 15 psf = 55 psf. Further, the joist, obviously, must carry itself. Sometimes the weight of the joist itself is `included’ in the weight of the floor (e.g., floor system). Or it may be examined separately. Often for the floor system, or volume, the estimated (or assumed) floor weight includes the joists, as the joists are part of the system (i.e., enclosed by floor sheathing above, gypsum ceiling below, and perhaps insulation between); but when examining, for example, a beam carrying the joists, the beam weight will be explicitly (additionally) examined. Back to our example.

The total load is 55 pounds per square feet. The area (square feet) supported by each joist is … 16 inches / 12 inches per foot times 10 feet equals 13.33 square feet (sf or sq ft). Thus the total load on the joist is … 55 psf times 13.33 sq ft = 733 lb (W = 733 lb).

Before we drop this number into our equation, we need E and I.

We generally `look up’ E. Perhaps the best place to get it is the National Design Specification® for Wood Construction (NDS) by the American Wood Council. In its Supplement are provided design values, including E values, for commonly used wood products. Let’s, for our example, use Douglas-fir / Larch Grade No. 2 joists. In the Supplement we find E = 1,600,000 psi.

(Strictly speaking, modulus of elasticity is psi per inch per inch, … the amount of stress required to produce a unit deflection … but the inch per inch is typically not shown.)

The moment of inertia, I, is a function of the joist cross section, and how oriented. By formula, for a rectangular cross section,

I = b h3 / 12,

where

b = dimension of the side, or edge, perpendicular to the load direction, and

d = dimension of the side parallel to the load direction.

For a floor, the load is downward; b is the horizontal top or bottom dimension, and d is the dimension of the vertical sides of the joist. Typically we orient the joist `on edge’, so b is the narrow edge, and d is the wider face, or depth.

For a dimension lumber 2 x 10, the actual dimensions are 1.5 in. x 9.25 in.

For our joist,

I = b h3 / 12 = 1.5 in. (9.25 in.)3 / 12 = 98.9 in.4.

Here we go …

The calculated deflection for the intended loading condition is … (10 ft is 120 inches) …

Δ = 5 W L3 / 384 E I = 5 (733 lb) (120 in.)3 / [ 384 (1,600,000 psi) (98.9 in.4) ] = 0.104 in.

A typical limitation on floor deflection is that the sag under live load be not greater than the span divided by 360, in this case, 120 inches / 360 or 0.33 inches.

Since the deflection calculated using the total load (live and dead) is 0.104, of which the deflection due to live load would be just a fraction, the deflection is acceptable.

But there’s more we need to talk about.

For starters, E is not a `true’ E. In the lumber world, published values for E, though often not obvious, are `apparent’ values. They arise from actually bending lumber pieces under load, in a `lab’, and back- calculating the E, using an equation(s) `like’ the one above. The problem is, the equation above, and others like it, are based on bending (beam curvature) arising solely from the tension and compression stresses in the lumber, where tension on the bottom half of the beam stretches the wood fibers, and compression on the top half of the beam compress the fibers, resulting in a curved shape. In real life, when a beam is loaded, it bends (sags) a bit more than that caused by the tension and compression stresses described above; as the wood also `shears’ (has shear distortion). The value obtained by this bending test, nevertheless, is `published’ … but, is usable for deflection calculations of other lumber, suffering the same phenomenon; that the resulting deflection includes the effect of `pure’ bending (tension and compression), and shear.