Comparison of Beam Stability Equations for a Wood Beam

Introduction

In the design of a wood beam we typically make sure the beam doesn't shear, doesn't rupture (break open), doesn't become unstable, and make sure that it doesn't flex so much as to become unserviceable. In this lesson we will look at the `unstable' part. Specifically, if the compression zone of a beam is not braced laterally, it might buckle (and twist) sideways, also known as lateral torsional bucking. We will investigate lateral torsional buckling of an example beam using the current design practice of the 2005 National Design Specification and compare (or contrast, if need be) our result to what is obtained using the procedure of the American Wood Council Technical Report 14, Designing for Lateral-Torsional Stability in Wood Members.

Example

Let's consider a Douglas fir 2400F-1.8E glued laminated timber beam of structural span 22 ft, subject to some Dead load and some Snow load. The beam is not supported laterally between the end supports. A trial size of 6-3/4 in. x 27 in. is to be investigated. The Dead load given above does not include the self weight of the beam. The loads are assumed to be applied uniformly along the beam.

Beam Stability in the National Design Specification

In The National Design Specification for Wood Construction (NDS) beam stability is addressed using the Beam Stability Factor, C_L, which is an adjustment factor used in determining allowable bending stress. The Factor is given by,

C_L = (1 + F_bE/F_b*)/1.9 - √{[(1 + F_bE/F_b*)/1.9]² - (F_bE/F_b*)/0.95},

where,

F_bE = the critical buckling design value for bending = 1.20 E_min' / R_B²,

where,

E_min' = adjusted modulus of elasticity for beam stability calculations = E_min multiplied by adjustment factors C_M, C_t,

R_B = beam slenderness ratio = √(l_e d / b²),

where,

l_e is the effective length of the beam, from Table 3.3.3 in the NDS,

b is the beam width, and d the depth,

and,

F_b* = the bending design value multiplied by all adjustment factors except C_L, C_V, and C_fu,

hence,

F_b* = F_b C_D C_M C_t C_C,

and, in no case may a beam with slenderness ratio greater than 50 be used.

For lateral torsional buckling the weak axis (y-y) value for E (E_min) is used. From the NDS Supplement this is found to be 830,000 psi (Table 5A).

Thus, for our example, assuming dry service conditions and normal temperatures,

E_min' = 830,000 (1.0)(1.0) = 830,000 psi.

The effective length in bending, l_e, is obtained from NDS Table 3.3.3 based on the unsupported length l_u and beam depth d.

From the Table, for a value of l_u/d = (22 ft × 12 in. / ft)/27 in. = 9.78, l_e = 1.63 l_u + 3d (for l_u/d ≥ 7) = 511 in.

Thus,

R_B = √[511 in. × 27 in. / (6.75 in.)²] = 17.4 (which does not exceed 50; good).

So,

F_bE = 1.20 E_min' / R_B² = 1.20 (830,000 psi) / (17.4)² = 3287 psi.

F_b* = Fb (1.15)(1.00)(1.00)(1.00, no curvature), where F_b = 2400 psi; so, Fb* = 2400 psi (1.15) = 2760 psi.

Finally, then,

C_L = (1 + 3287/2760)/1.9 - √{[(1 + 3287/2760)/1.9]² - (3287/2760)/0.95} = 0.877.

Beam Stability using AWC TR 14

From Section 2.1.3.1 of TR 14,

C_L = (1 + α_b)/1.9 - √{[(1 + α_b)/1.9]² - (α_b)/0.95},

where,

α_b = M_cr/M*,

where

M_cr = critical buckling moment = 1.3 C_b C_e E_{y 05}' I_y / l_u,

where,

C_b = equivalent moment factor,

C_e = load eccentricity factor,

E_{y 05}' = adjusted weak axis fifth percentile modulus of elasticity (fifth percentile E adjusted free of shear and divided by 1.66 factor of safety),

I_y = weak axis moment of inertia,

and,

l_u = the unbraced length.

E_{y 05}' = the same as E_min' in the 2005 NDS approach = 830,000 psi in our example.

C_b may be taken from TR 14 Table 2, which for our example is ... 1.13 (uniformly distributed load without lateral support).

C_^e = √(η² + 1) - η (but need not be less than 0.27),

where,

η may be taken to be 1.3 k d / l_u for rectangular members, and where k may also be obtained from TR 14 Table 2; k = 1.44 for our example.

So,

η = 1.3 (1.44) (27 in.) / (22 × 12 in.) = 0.191.

And,

C_^e = √(0.191 ² + 1) - 0.191 = 0.827.

I_y for our beam is 692 in.⁴ (calculated or per Table 1C in the NDS Supplement).

Thus,

M_cr = 1.3 (1.13)(0.827)(830,000 psi)(692 in.⁴)/(22 × 12 in.) = 2,643,000 lb-in. = 220,300 lb-ft.

And,

M* = Moment resistance in strong axis bending multiplied by all applicable adjustment factors except C_fu, C_V, and C_L = F_b S_x C_D C_M C_t C_C = 2400 psi (820 in.³) (1.15)(1.00)(1.00)(1.00) = 2,263,000 lb-in. = 188,600 lb-ft (where S_x may also be calculated or obtained From the NDS Supplement).

So,

α_b = 2643 / 2263 = 1.168,

thus,

CL = (1 + 1.168)/1.9 - √{[(1 + 1.168)/1.9]² - (1.168)/0.95} = 0.872.

Discussion

The C_L values using the two methods compare nicely. Good.

Before we close, however, let's calculate C_V, as C_L doesn't even apply if C_V is less than C_L.

For western species softwoods,

C_V = (5.125 in. / b)^1/10(12 in. / d)^1/10(21 ft /L)^1/10, which in our case gives us,

C_V = 0.893 (L = 22 ft).

So, C_L controls, but not by much.

References

National Design Specification for Wood Construction and Supplement Design Values for Wood Construction, 2005 Edition, American Forest & Paper Association / American Wood Council, 1111 Nineteenth St., NW, Suite 800, Washington, D.C., 20036, www.awc.org.

Designing for Lateral Torsional Stability in Wood Members, Technical Report 14, American Forest & Paper Association / American Wood Council, 1111 19th St., NW, Suite 800, Washington, DC 20036.