The Beam Stability factor accounts for the possibility of a beam buckling (rotating laterally) due to the compressive action in the concave portion(s) of the beam. In a great number of timber beam applications the compression zone is typically held in place against this lateral-torsion buckling by deliberate attachment of the system that the beam is supporting (floor, roof, etc.). In some cases, however, particularly with exposed, continuous beams where the compression zone is on the bottom face and is not supported laterally, potential buckling must be considered. The `standard’ way of dealing with this potential buckling is to determine a `Beam Stability factor’ which is used as an Adjustment factor on the Bending Design Value. The Beam Stability factor reduces (reserves) the potential flexural capacity of the beam to prevent such buckling. Other common examples for construction for which the Beam Stability factor must be considered are for beams for which the compression zone is on the top of the beam but for which the top of the beam is not or only partially supported laterally, such as beams supporting walls or columns but without the lateral support of a floor, and beams supporting secondary girders or other beams, leaving unsupported segments of the beam.

In this example we will engage the somewhat cumbersome determination of the Beam Stability factor for a continuous timber deck beam unsupported along its bottom face except at the very ends.

The common approach for determining the Beam Stability factor for a wood bending member is the `Equivalent Length Method’. However, the proper application of this Method to continuous beams is a bit obscure. In its stead we will use in this example the `Equivalent Moment Method’ commonly used in steel beam design. Presentation of this method for Glulam design is presented in the Sixth Edition *Timber Construction Manual* (TCM) and mirrored for structural wood in general in Technical Report No. 14, *Designing for Lateral Torsion Bucking in Wood Members* (TR 14), American Forest & Paper Association / American Wood Council.

Using the Equivalent Moment Method,

CL = (1 + α )/1.9 – √ { [ (1 + α)/1.9 ]2 – α / 0.95 },

where

CL is the Beam Stability factor, and

α is the ratio FbE / Fb*, where

FbE is the critical buckling design value for bending members, and

Fb* is the Bending Design value multiplied by all applicable Adjustment factors except CL (obviously), CV, and Cfu.

FBe = 1.20 Emin’ / RB2,

where

Emin’ is the adjusted `minimum’ Modulus of Elasticity with respect to the direction of potential lateral buckling, and

RB is the Slenderness Ratio of the beam;

RB = √[(1.84 *lu* d)/(Cb Ce b2)],

where

*lu* is the unsupported length,

d is the beam depth,

Cb is the Equivalent Moment Factor,

Ce is the Load Eccentricity Factor, and

b is the beam width.

The unsupported length *lu* should be taken to be the distance between points along the beam for which the tension and compression zones are prevented from relative rotation. Load points may or may not provide this resistance; if not, the entire beam length should be used (assuming the ends are held in place against relative rotation, *which they better be!*)*.*

Ce is 1.0 for cases where the beam is braced at the load point, or where the load is applied to the tension face or through the neutral axis. For other cases,

Ce = √(η2 + 1) – η,

where

η = 1.3 k d / *lu*,

where

k is obtained in tabular form for simple and cantilever beams and may be taken (conservatively) to equal 1.72 for other cases.

Cb may be taken to equal 1.0 for all cases or may be calculated;

Cb = 12.5 Mmax / (3 MA + 4 MB + 3 MC + 2.5 Mmax)

where

Mmax is the absolute maximum moment between the points of bracing,

MA is the absolute moment at *l*/4,

MB is the absolute moment at *l*/2,

MC is the absolute moment at 3*l*/4,

where

*l* is the distance between locations braced against rotation.

The particular beam under investigation is a 5-1/8 x 10.5 24F-1.8E Douglas fir beam 36 ft long with three equal spans of 11.33 ft and 1 ft overhangs at each end. The beam is braced against rotation at the end supports only. The beam is loaded uniformly with a total load of 805 pounds per linear foot (plf) which includes Snow and Dead loads and beam-self weight.

For simplicity (or at least initially) conservative values of (…)

Cb = 1.0 (to avoid determining the moments at the quarter and mid-points),

and k = 1.72 (for the calculation of Ce).

The unbraced length, *lu,* is 36 – 1 – 1 = 34 ft = 408 in.

For the 24F-1.8E beam,

Fb = 2400 psi,

E min’ = E min y’ = E min y = 850,000 psi (no applicable adjustments; weak y-axis bending for buckling),

and

CD = 1.15 (Snow load duration),

b = 5.125 in., and d = 10.5 in.

η = 1.3 (1.72)(10.5 in.)/408 in. = 0.0575.

Ce = √(0.05752 + 1) – 0.0575 = 0.944.

RB = √[(1.84 x 408 in. x 10.5 in.)]/(1.0 x 0.944 x 5.1252 in.2)] = 17.8.

In no case may RB exceed 50; good!

Fb* = 2400 psi (1.15) = 2760 psi.

FbE = 1.20 (850,000 psi)/(17.8)2 = 3219 psi.

α = 3219/2760 = 1.17.

Thus,

CL = (1 + 1.17)/1.9 – √{[(1 + 1.17)/1.9]2 – 1.17/0.95} = 1.14 – √{1.142 – 1.23] = 0.88.

In a previous article a Beam Stability factor *estimate* of 0.90 was used for this particular example. The beam was found to have plenty of capacity with respect to this estimate, so an exact Stability factor was not calculated. A less conservative approach, if needed, could be to determine Cb from the equation above (repeated below),

Cb = 12.5 Mmax / (3 MA + 4 MB + 3 MC + 2.5 Mmax).

A detailed analysis (not shown here) yields MA and MC = 24,700 lb-in., MB = 33,000 lb-in., and M max = 122,000 lb-in. (absolute values).

Thus,

Cb = 12.5 (122) / (3 x 24.7 + 4 x 33 + 3 x 24.7 + 1.5 x 122) = 2.6.

Plugging this into RB,

RB = √[(1.84 x 408 in. x 10.5 in.)]/(2.6 x 0.944 x 5.1252 in.2)] = 11.

Thus,

FbE = 1.2 (850,000) / 112 = 8340.

α = 8340/2760 = 3.0.

Thus,

CL = (1 + 3)/1.9 – √{[(1 + 3)/1.9]2 – 3/0.95} = 0.98.

This `more exact’ calculation of the Beam Stability factor gets us much closer to Unity, as we might expect for the relatively stout beam with relatively short individual spans. For the uniformly loaded three-equal-span continuous beam the quarter, three-quarter, and mid-span moments are much smaller values relative to the maximum, driving Cb up and thus CL.

Some designers might use the length of beam between inflection points (length of beam in negative bending) as the unbraced length, or the `effective’ length in the Effective Length method. Some discussion of this subject is in the new Sixth Edition of the *Timber Construction Manual*. The actual length of beam in negative bending for this example is quite short, less than 0.5*l*, which would also drive CL to near Unity.

References

Design of a Glued Laminated Timber Deck Beam, Jeff Filler, Associated Content.

*Timber Construction Manual*, Sixth Edition, American Institute of Timber Construction, Centennial, CO.

Technical Report No. 14, *Designing for Lateral Torsion Bucking in Wood Members* (TR 14), American Forest & Paper Association / American Wood Council, Washington, DC.

*Standard Specification for Structural Glued Laminated Timber*, AITC 117, 2004 (and 2010), American Institute of Timber Construction, Centennial, CO.