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 L²/8,

Shear, V = 5 w L / 8, and

Deflection, Δ = w L⁴ / 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 ½ 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,

Δ = (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.