Beam Stability Factor Using 2005 Dual Format NDS

Consider a 5-1/8 in. x 24 in. 24F-1.8E Douglas fir glulam beam subject to the following loads: 325 plf Dead load and 1060 plf Occupancy Live load. The beam is `simple' and spans 20 ft - 4 in. (20.33 ft) The stated loads do not include the weight of the beam itself. Let's do the flexure (bending) check for the beam using both the ASD and LRFD methods as presented in the 2005 Dual Format National Design Specification ®for Wood Construction (NDS). Assume the beam is held in place and prevented from rotation at the ends, but is not braced along its span to resist lateral torsional buckling. (Consider dry and normal temperature service conditions.)

Since the beam is not braced (in particular the top / compression zone of the beam), the Beam Stability Factor, C_L, must be considered.

First let's determine the self weight of the beam, since it must be added to the Dead load. Using a specific weight of 33 pcf (Timber Construction Manual, Table 2.2),

ω _{self weight} = (5.125/12 ft)(24/12 ft)(33 pcf) = 28 plf.

Thus the total Dead load for the beam is,

ω _{applied Dead + self weight} = 325 + 28 = 353 plf.

ASD

Using the ASD approach the loads combine as follows:

D + L = 353 plf + 1060 plf = 1413 plf Total load ...

The maximum bending moment in the beam is at mid-span and is,

M _max = ω L² / 8 = 1413 plf (20.33 ft)² /8 = 73,000 lb-ft = 876,000 lb-in.

The Section Modulus for the beam is,

S = bh²/6 = (5.125)(24)²/6 = 492 in.³ (or by Timber Construction Manual Table A.1.2.1, or NDS® Table 1C).

The extreme fiber bending stress is thus,

f_b = M/S = 876,000 lb-in./492 in.³ = 1780 psi.

This value is checked against the allowable bending stress, which is obtained by multiplying the reference design value for bending by all the appropriate adjustment factors. The allowable stress is,

F_b' = F_b C_D C_M C_t (the lesser of C_L and C_V) ...

Where, for this problem,

C_D = 1.00 (Normal Load Duration for Occupancy Live load), C_M = 1.00, C_t = 1.00, and C_L and C_V will be calculated. (You guessed it: `M' for moisture and `t' for `temp'.)

First, F_b = 2400 psi.

And note: in the 2005 Dual Format NDS, this F_b=2400 psi is for both methods.

C_V is the Volume factor and is calculated as follows (NDS® Section 5.3.6):

C_V = (5.125 in./b)^1/x(12 in./d)^1/x(21 ft/L)^1/x ,

where x = 10 for Douglas fir,

thus,

C_V = (5.125/5.125)^1/10(12/24) ^1/10 (21/20.33) ^1/10 = 0.936.

Now for the Beam Stability Factor, C_L ...

The Beam Stability Factor is more grueling.

The equation for the Beam Stability Factor in the NDS ® is the same for both ASD and LRFD. From NDS ® Section 3.3 ...

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

It is in the determination of F_bE and F_b* that they (the two methods) differ.

For both methods,

F_bE = 1.20 E_min' / R_B².

In ASD,

E_min' = E_min C_M C_t,

where

E _min is the Reference Modulus of Elasticity for Beam and Stability Calculations ... the same number ... to be obtained from NDS ® Supplements, whether using ASD or LRFD;

and where C_M, C_t are the applicable adjustment factors, using ASD, from Table 5.3.1 in the NDS®, in this case for moisture and temperature, which in this particular example, the first two are unity.

Thus we obtain E _min from the NDS® Supplement; from Table 5A, using E _{y min} since we are looking at the beam buckling in the `weak' direction, we get,

E _min = E _{y min} = 830,000 psi.

So, then, in our example,

E _min ' = 830,000 psi (1.0)(1.0) = 830,000 psi (ASD).

R_B = √(l_e d / b²), called the slenderness ratio of the bending member, and not permitted to exceed 50.

where,

l_e is the effective span length from NDS® Table 3.3.3,

which,

for l_u / d = 20.33(12)in./ 24in. =10.1 ≥ 7 is,

(l_e = ) 1.63 l_u + 3 d = 1.63(20.33)(12) + 3(24) = 469.7 in. (same for both ASD and LRFD),

... l_u being the unsupported length (greatest unsupported length of the top/compression zone) of the beam.

So,

R_B= √(l_e d / b²) = √[(469.7 in.)(24 in.)/(5.125 in.)²]= √(429) = 20.7 (same for both ASD and LRFD).

Baby!

So, now,

F_bE= 1.20 E_min' / R_B² = 1.2 (830,000 psi)/(20.7)² = 2321 psi ( ... ASD).

Now to deal with Fb*.

We start at the same F_b for both formats.

NDS® Section 3.3 defines Fb* as the reference bending design value multiplied by all applicable adjustment factors except C_fu, C_V and C_L.

Thus, for ASD, from Table 5.3.1,

F_b* = F_b C_M C_t C_c, (ASD)

the first two being unity, and likewise the third (or not applicable), as it is for curved beams.

So,

F_b* = 2400 psi (1.0)(1.0) = 2400 psi. (ASD)

To make the number crunching a bit easier,

F_bE/F_b* = 2321/2400 = 0.967. (ASD)

So,

C_L = (1 + 0.967)/1.9 - √{[(1 + 0.967)/1.9]²-(0.967)/0.95 = 0.803. (ASD)

This is less than C_V, so C_L controls.

So, now, the allowable stress for bending is,

F_b' = F_b C_M C_t C_L = F_b* C_L = 2400 psi (0.803) = 1927 psi. (ASD)

The design check becomes,

... f_b = 1780 psi ≤ F_b' = 1927 psi; good. (ASD)

Or, in terms of a unity check,

... f_b / F_b' = 1780 psi / 1927 psi = 0.924 ≤ 1.00; good. (ASD)

LRFD

For LRFD we must first determine the factored load.

From ASCE 7, Section 2.3.2, the Dead and Occupancy Live loads combine as,

1.2 D + 1.6 L.

So, in our case,

... ω = ω_u = 1.2 (353 plf) + 1.6 (1060 plf) = 2120 plf. (LRFD)

The maximum bending moment in the beam is at mid-span and is thus,

M _max = ω L² / 8 = 2120 plf (20.33 ft)² /8 = 109,500 lb-ft = 1,314,000 lb-in. (LRFD)

The Section Modulus for the beam is the same.

The extreme fiber bending stress is thus,

f_b = M/S = 1,314,000 lb-in./492 in.³ = 2671 psi. (LRFD)

This value is checked against the reference design value in bending multiplied by all appropriate factors including the factors associated with LRFD;

F_b' = F_b C_M C_t (the lesser of C_L and C_V) C_fu C_c λ K_F ϕ_b ... (NDS Table 5.3.1 for glulam)

Where, for this problem,

C_M = 1.00, C_t = 1.00 from before; C_V is = 0.936, the same as before. C_L must be recalculated per the LRFD factors, and the lesser of it and C_V used.

F_b = 2400 psi as before.

F_bE = 1.20 E_min' / R_B².

In LRFD,

E_min' = E_min C_M C_t K_F ϕ_s

where

E _min is the same as before = E _{y min} = 830,000 psi.

K_F for E min is obtained from Appendix Table N1, along with the K_F for bending;

K_F for E min is 1.5 / ϕ,

K_F for bending is 2.16 / ϕ.

For the ϕ values; Table N2,

ϕ for bending is ϕ_b = 0.85,

ϕ for E _min is ϕ_s = 0.85.

So, E _min' = 830,000 psi (1.0)(1.0)(1.5/0.85)(0.85) = 1,245,000 psi. (LRFD)

R_B= 20.7 (same for both ASD and LRFD).

So,

F_bE= 1.20 E_min' / R_B² = 1.20 (1,245,00 psi)/(20.7)² = 3481 psi. ( ... LRFD)

Now for Fb*.

For LRFD, from Table 5.3.1,

F_b* = F_b C_M C_t C_c K_Fϕ_b λ (LRFD)

λ = 0.8 from NDS® Table N3.

So,

F_b* = 2400 psi (1.0)(1.0)(1.0)(1.0)(2.16/0.85)(0.85)(0.8)= 4147 psi. (LRFD)

Again, to make crunching C_L a bit easier,

F_bE/F_b* = 3481/4147 = 0.839.

So,

C_L = (1 + 0.839)/1.9 - √{[(1 + 0.839)/1.9]²-(0.839)/0.95 = 0.736. (LRFD)

This is less than C_V, so C_L controls.

So, now,

F_b' = F_b* C_L = 4147 psi (0.736) = 3052 psi. (LRFD)

The design check becomes,

... f_b = 2671 psi ≤ F_b' = 3052 psi; good. (LRFD)

Or, in terms of a unity check,

... f_b / F_b' = 2671 psi / 3052 psi = 0.875 ≤ 1.00; good. (LRFD)

References

National Design Specification (NDS) ®for Wood Construction, 2005, and Supplements, American Forest & Paper Association, Washington, D.C.

Timber Construction Manual, 5^th Edition, American Institute of Timber Construction, Centennial Colorado.

Minimum Design Loads for Buildings and Other Structures, ASCE 7 (ASCE 7-05), American Society of Civil Engineers.