E app (Apparent Modulus of Elasticity)

E app (`Eee app’, `Eee apparent’, Apparent Modulus of Elasticity) is a design property for wood construction. It comes from the Modulus of Elasticity of the wood (E, MOE, Young’s Modulus). It’s not a `true’, or `pure’ property, and it is not the same as `True E’ (or True Modulus of Elasticity, or E true). It comes about from actually bending beams to determine E, to be used in the design of beams (joists, rafters, girders, other spanning members). The background is this: beam deflection theory (an article itself) gives us equations (formulas) that we use to calculate the deflection (bend, bow, sag) of beams under load. Determining deflections is important as we don’t want beams to sag `too much’ … to where they might look, or feel unsafe, or otherwise cause problems. These equations are based on the flexural stresses in the wood. In a simple beam under downward load, the bottom of the beam is stretched, and the top of the beam is compressed, resulting in a curved or bowed downward shape. The equation used, for example, to determine the amount of deflection (`bow’) due to a concentrated load at the mid-span of a so-called `simple’ beam is …

Δ = P L3 / (48 E I),

where

Δ = the amount of deflection, or sag,

P = the amount of load (e.g., the weight of a person, or downward load delivered, say, from post or column)

L = beam length

48 = a `number’, based on the distribution of the load and the support conditions of the beam (in this case the beam is supported at the ends, the ends are allowed to `rotate’, and the load is concentrated at the mid-span)

E = the Modulus of Elasticity, or Young’s Modulus (E), and

I = the Area Moment of Inertia of the beam shake (cross section).

Let’s take, for example, the deflection caused by a person weighing 185 pounds (lb) standing mid-span on a 2 x 12 Douglas-fir piece of dimension lumber laid flat (as a plank), supported at ends by saw horses 8 feet (ft) apart.

Thus,

P = 185 lb and L = 8 ft or 96 inches …

E = where to get? … Let’s see … we could look in our Mechanics of Materials textbook(s) from college, e.g., Gere, 5e, p 899 … shows, for Douglas fir … a range of 1,600 to 1,900 ksi. Hmmm, that’s not going to be very precise! (Of course, ksi stands for 1000s of psi, and psi stands for pounds per square inch.) Let’s try and be more exact. Wood is a highly variable building material. Wood is variable between species, and within species. To pin down E, let’s look to some resources more specific to wood. One source is the Wood Handbook, Wood as an Engineering Material, United States Department of Agriculture, Forest Service, Forest Products Laboratory … my hard copy says printed by the Forest Products Society, 1999. There’s a newer version available online (free). (I suggest you get it if you are at all interested in engineered wood! I consider it the Bible of engineered wood.) Page 4-12, Douglas-fir … we’ll pick `Inland West’1.83 x 106 psi. Let’s also cruise over to the 2018 American Wood Council (AWC) National Design Specification® for Wood Construction (NDS). Table 4A (in the NDS Supplement) gives us, for E, … 1,800,000 psi, for Grade No. 1 & BTR … 1,700,000 psi for No. 1, 1,600,000 psi for Grade No. 2, and so on. Which one do we pick? Pick the one that you intend to use, or already have on hand, or are willing to buy. (Or the one you end up requiring for your design situation.)

Back to I … the (Area) Moment of Inertia of the beam (plank) cross section.

For a rectangular shape,

I = b h3 / 12,

where

b = the beam (plank) width, and

h = the depth.

The actual dimensions of a so-called `2 x 12’ are 1.5 in. x 11.25 inches.

Since the 2 x 12 is laid flat, b becomes 11.25, and h = 1.5; so …

I = (11.25 in.) (1.5 in.)3 / 12 = 3.16 in.4.

For 2 x 12 Douglas-fir Grade 1 & BTR (E = 1,800,000 psi),

Δ = P L3 / (48 E I) = 185 lb (96 in.)3 / (48 x 1,800,000 psi x 3.16 in.4) = 0.599 in. (about 5/8th of an inch).

I’ll bet we can see that! I’ll be you can also feel that if you happen to be the 185-lb load standing mid-span on the plank.

Now, we need to look at some fine print. If we look at the NDS, i.e., Appendix F, it says, first of all, that the `E’s found in Table 4A are average values, and that individual pieces will have higher and lower values. So, don’t be surprised if you actually conduct the plank scenario above, and get a value different than 0.599 in. In fact, you should expect a different value, more or less. Second, Appendix F3 tells us that the values include a `shear deflection component’. What is that? Well, when we bend wood, the wood fibers that are compressed shorten, the ones that are stretched lengthen, resulting in a curved shape, but the wood also `shears’ (rotates). Look at Gere p. 345 (stress) and p. 525 (strain). If we were to draw a perfect square on the side of the beam (mid-depth), it would shear (rotate) and become slightly `not square’, but a trapezoid. This `shearing’ causes the beam to deflect a bit more than just from the pure compressing and stretching of the wood fibers. If we read on, F3.1 says that the ratio of the shear-free E to the reference (table value) E is 1.03 (for sawn lumber). So, the published E values are ~ 3% less than true E (Young’s Modulus E, etc.), resulting in slightly more deflection (due to shear) than if we had used true E (Young’s Modulus) alone in the above equation.

If we backtrack, we’ll read on p. 4-2 of the Wood Handbook, that the value we got from 4-12 was from static bending and included (includes) the effect of shear. Further, it is stated that this effect can be approximately removed by increasing the table value by 10%. Footnote c to the table from which we plucked 1.83 (million psi) says the same thing. Backtracking further, we might have seen our first hint in all this in Gere, where, next to wood, for Modulus of Elasticity, it shows `(bending)’.One thing is certain: shear stresses in a wood beam increase the beam deflection. We could also say, with certainty, that an `apparent Modulus of Elasticity’, which is obtained by bending a beam, and back-calculating a modulus of elasticity using the `pure bending’ or `beam deflection formula(s)’, is less than the `true’ E (relationship between tension and compression forces, and resulting deformations). How did all this come about? Well, it’s more convenient, and relevant, to come up with a number for E for wood, using a bending test, versus some kind of compression or tension test. This is not necessarily a bad thing, either, since the huge bulk of engineering calculations performed on wood are indeed for its use in some kind of `bending’ application. (And accompanying shear stresses are essentially unavoidable in typical bending applications.) How much a beam actually bends (deflects, `bows’) is a big deal … an important design consideration. How much a wood post compresses under load, on the other hand, and based on a true E (without a shear effect), is rarely examined.

Several more things: 1) the current trend in the design of, especially, engineered wood products (EWP), e.g., prefabricated wood I-joists, glued-laminated timber, structural composite lumber (SCL), is to more deliberately separate (and perhaps publish both) the E app and E true values. We’ll talk more about this on another post; 2) formulas are available for explicitly determining the component of deflection due to shear stresses. The deflection equation thus takes on an additional term; the first term looks like the one we used above, accounts for the deflection due to elongation and shortening of the wood fibers, via Young’s Modulus, and uses the true E, and a second term that calculates the deflection due to shearing. Ironically, both components of the equation typically require true E for input. Theoretically, shear deformations/deflections are based on G, the modulus of rigidity, which is hard to pin down for wood. It is often taken as a function of (fraction of) E.

One thing that remains `not quite as certain’, at this point in this conversation, is the exact relationship (ratio) between `apparent’ E and `true’. Is E app 3% less (from the NDS, for sawn lumber)? … or E without shear 10% more (from the Wood Handbook)? Since the NDS is the `Standard’ used for wood design and construction, I’ll stick with the NDS value.NDS Appendix F3 indicates ratios of shear-free E to reference E of 1.03 for sawn lumber, and 1.05 for glued laminated timber. What about SCL (structural composite lumber)? Prefab wood I-joists? We’ll examine these more in other posts. But, for now, all of this is almost becomes `academic’, contrast to the variability of wood, and its properties such as Modulus of Elasticity, which we’ll also discuss in another post.

There is another reason we might need E true, however, for a wood beam, and that’s for the determination of E min, which is derived from true E, and used for stability analysis. We’ll save E min for another post.