# Step-by-Step Design Calculations for a Wood Beam for a Deck

In a previous article we tackled the design of the framing elements (decking, joists, support beams, and posts) for a 12′ x 32′ exterior deck for a mountain cabin. In that design we came up with four 3-ply 2 x 12 Douglas Fir – Larch (DF) beams, each 8 feet (ft) long; the inner beams spanning 8 ft `simply’, and the outer beams 7 ft with 1 ft overhangs. In this article we examine the use of the same 2 x 12s, except that instead of four 8-ft beams, we will use two 16-ft beams. These two 16-ft beams will span 8 ft, 7 ft, and then over the outer supports to overhang 1 ft. The advantage to using two longer beams (instead of four) will be better framing `continuity’ and reduced beam deflection (sag).

The original design using the 8-ft simple spanning beams was `easy’; we used a `Wood Beam Capacity Calculator’ that we *found* on line. That Calculator, however, does not handle continuously spanning beams. This article covers the more cumbersome `hand calculations’ that we will need instead.

We will start by determining the load on the beam, in pounds per linear foot (plf). Then we will calculate the `internal forces’ (Bending Moment and Shear) and the beam deflections. From the internal forces we will calculate the internal *stresses* (bending stress and shear stress). Then we will check the stresses due to load with regard to the `Allowable’ stresses for the beams (based on species, grade, and service conditions). And we will make sure the deflection under load is not excessive. We will start with the same 3-ply 2 x 12 built-up beam, Select Structural grade lumber, and then add or subtract plies, change lumber size, if necessary, until everything `checks out’.

To get the internal forces we will model each 8 + 7 + 1 ft beam as having two 8 ft continuous spans. Item 29a of Table B.1 from the *Timber Construction Manual* gives the following:

Bending Moment, M = w L^{2}/8,

Shear, V = 5 w L / 8, and

Deflection, ÃŽÂ” = w L^{4} / 185 EI,

where

w = line load on the beam,

L = span length (support to support),

E = Modulus of Elasticity of the wood, and

I = Moment of Inertia of the beam section.

The beams will carry approximately half of the 11 ft joist spans plus the 1 ft joist overhangs, for a tributary width of Ã‚Â½ of 11 + 1 = 6.5 ft. Thus, the line load on the breams will be,

w = 260 psf x 6.5 ft = __1690 plf.__

Let’s add 10 plf for the weight of the beams themselves, or

w (total carried by beams) = (total carried by beams) = __1700 plf.__

From the *National Design Specification for Wood Construction – Supplement – Design Values for Wood Construction*, we obtain the following Design Values and section properties for Douglas Fir Select Structural Dimension Lumber:

F_{b} = 1500 psi,

E = 1,900,000 psi, and

F_{v} = 180 psi. (Yes, we will also check Shear.)

We will be using the 2 x 12s on edge (strong x-x direction); the section properties for a single 2 x 12 are:

Area, A = 16.88 in.^{2},

Section Modulus, S = 31.64 in.^{3}, and

Moment of Inertia, I = 178 in.^{4}.

For a `3 ply’ we will have:

__A = 3 x 16.88 = 50.6 in. ^{2}__,

__S = 3 x 31.64 = 95 in. ^{3}__, and

__I = 178 x 3 = 534 in. ^{4}__.

The bending moment is,

M = 1700 plf (8 ft)^{2} / 8 = 13,600 lb-ft = 163,200 lb-in.

The bending *stress*, f_{b}, is,

f_{b} = M/S = 163,200 lb-in. / 95 in.^{3} = __1718 psi__.

The `allowable’ bending stress (F_{b}‘) is the F_{b} multiplied by appropriate adjustment factors;

F_{b}‘ = F_{b} (from the Supplement) x 1.15 (Load Duration factor for Snow load) x 1.0 Size Factor (see Supplement) x (0.9 presumed Stability factor) x 1.15 (Repetitive Member factor, 3 or more members side-by-side, see Supplement), or

F_{b}‘ = 1500 psi x 1.15 x 0.9 x 1.15 = __1785 psi__.

Since the bending stress is 1718 psi, the 3 ply 2 x 12 is strong enough with regard to bending stress.

Now we check Shear in the beam.

The Shear *stress* in the (rectangular shape) beam is,

f_{v} = 3 V / 2 A,

where

V = 5 w L / 8 = 5 (1700 plf)(8 ft) / 8 = __8500 lb__.

Thus,

f_{v} = 3 (8500 lb) / [2 x 50.6 in.^{2}] = __252 psi__.

The `allowable’ shear stress is the published value from the Supplement multiplied by the appropriate adjustment factors for shear,

F_{v}‘ = 180 psi x 1.15 (for Snow) = __207 psi__.

This is a bit scary in that the design stress ends up being more than the allowable stress. Not good! However, let’s check one thing. The *National Design Specification* (NDS) allows `uniform’ loads within a distance `d’ of the supports to be omitted in the shear design check, where d is the depth of the member. In our case, then, we may deduct 11.25 twelfths of the 1700 plf for V. Expressed in equation form,

V (design) = V (@ support) – w d, or,

V (design) = 8500 lb – 1700 plf (11.25/12 ft) = 8500 lb – 1594 lb = __6906 lb.__

The correspond *stress* is,

f_{v} = 3 (6906 lb) / [2 x 50.6 in.^{2}] = __205 psi__.

Sweet! … we did it! The stress under load, 205 psi, does not exceed the allowable, 207. (CLOSE!)

The deflection under the full load will be,

ÃŽÂ” = (1700 / 12 pounds per inch) (8 x 12 in.)^{4} / [185 x 1,900,000 psi x 534 in.^{4}] = __0.064 in.__ ( … about 1/16^{th} of an inch).

With respect to the span, this corresponds to 0.06 in. / (8 x 12 in.) = 0.00067 or about 1/1500.

Going back to our deflection limit of `span / 240’for a deck covered with snow (original article), we can see that the 3-ply 2 x 12 beams will be plenty stiff enough.

The 3 – ply 2 x 12 built-up beams WORK!!!

References

Structural Design of a Deck for a Mountain Cabin (Submitted).

Wood Beam Capacity Calculator by J. Ochshorn, Cornell University, Ithaca, New York.

*Timber Construction Manual* , 6^{th} Edition, John Wiley and Sons, 2012, Hoboken, New Jersey.

*National Design Specification for Wood Construction* and *Supplement – Design Values for Wood Construction*, 2005, American Wood Council, Washington, D.C.