Wood Column Design Using the 2005 NDS LRFD Provisions
Use of the NDS Column Stability Factor Using the 2005 NDS
FIRST LRFD
The axial load capacity will be determined by the setting the equation fc = P/A equal to Fc', where
fc = actual (applied) compression stress parallel to grain (under design loads),
A = cross sectional area of the column, and
Fc' = adjusted compression value parallel to grain (NDS® Section 1.6), and where
Fc' = Fc multiplied by all appropriate factors, where
Fc = the reference compression design value parallel to grain;
and where the appropriate factors are listed in NDS® Table 5.3.1 (glulam).
For LRFD these factors are (shown applied to Fc):
Fc' = Fc CM Ct CP KF ϕc λ,
where
CM and Ct = Wet Service and Temperature factors, in this example taken to be unity,
CP = the Column Stability Factor, which we will have to calculate,
KF = Format Conversion factor, for use with LRFD,
ϕc = Resistance Factor (a `strength reduction' factor) for compression, for use with LRFD, and
λ = the Time Effect Factor, for use with LRFD.
Here is CP (NDS® Section 3.7):
CP = (1 + FcE/Fc*)/2c - √{[(1 + FcE/Fc*)/2c]2 - (FcE/Fc*)/c}, (note: same equation for both ASD and LRFD)
where
FcE = critical buckling design value for compression members,
Fc* = reference compression design value parallel to grain multiplied by all applicable factors except CP,
and
c = value based on whether sawn lumber (0.8), round timber pole or pile (0.85), or glulam (0.9).
Let's tackle FcE;
FcE = 0.822 E min' / (le/d)2, where (note: same equation for ASD and LRFD),
E min' = adjusted modulus of elasticity for column stability calculations,
le = effective column length, equal to Ke l, where
Ke is our familiar buckling length coefficient for compression (yeah, Euler Buckling), in our case we'll use 1.0;
and
d is the `depth' of the member.
For a column that is unbraced in either/both directions, we need to examine E min' and le/d for both/each.
The direction that produces the lesser FcE will control CP and thus the design of the column.
From Table 5B of the NDS® Supplement for the DF No. 2 combination, we obtain E min = 830,000 psi. This value is applicable to both `directions' of buckling. We also get Fc = 1950 psi (4 or more laminations).
With E min and le the same for each direction, the lesser `d' will govern; thus
d = 5.125 in.,
le/d = 1.0 (9 x 12 in.)/ 5.125 in. = 21.07 (a slenderness ratio, which may not exceed 50, Section 3.7.1.4, and which is the same for ASD and LRFD),
and, for E min' we take E min and multiply it by the appropriate adjustment factors (Table 5.3.1):
E min' = E min CM Ct KF ϕs ( ... LRFD specific).
For the LRFD `factors', we go to NDS® Appendix N;
KF for E min is 1.5/ϕ,
ϕ for E min is 0.85,
and since we'll need them shortly,
KF for Fc is 2.16/ϕ,
ϕ for Fc is 0.90.
So,
E min' = 830,000 psi (1.0)(1.0)(1.5/0.9)(0.9) = 1,245,000 psi (LRFD specific).
And now,
FcE = 0.822 (1,245,000 psi) / (21.07)2 = 2305 psi. (LRFD specific)
Now we are almost ready to calculate CP.
For Fc* we go back to Table 5.3.1 (and get all the factors except CP);
Fc* = Fc CM Ct KF ϕc λ.
We need λ.
From Table N.3.3 we get λ = 0.8 for Dead plus Snow.
Actually, from the Table we also get the appropriate LRFD `Load Combination',
1.2 D + 1.6 S, (LRFD)
which we will use in a bit; otherwise we need to get the Load Combination from ASCE 7 (or IBC Section 1605).
SO,
Fc* = 1950 psi (1.0)(1.0)(2.16/0.9)(0.9)(0.8) = 3370 psi. (LRFD)
To make number crunching a tiny bit easier,
FcE/Fc*= 2305 psi / 3370 psi = 0.684.
So,
CP = (1 + 0.684)/2(0.9) - √{[(1 + 0.684)/2(0.9)]2 - 0.684/0.9} = 0.596. (LRFD)
And,
Fc' = Fc CM Ct CP KF ϕc λ = Fc* CP = 3370 psi (0.596) = 2009 psi. (LRFD)
The `capacity' of the column is thus (by setting fc = Fc'),
Fc' A = 2009 psi (5.125 in. x 6 in.) = 2009 psi (30.75 in.2) = 61,800 lb. (LRFD)
Recalling that in this example the loads divide out half Dead and half Snow,
Fc' A = 1.2 D + 1.6 S = 1.2 (0.5 x P total service load) + 1.6 (0.5 x P total service load) = 61,800 lb,
P total service load = 44,100 lb. (LRFD)
(Approx 22k Dead and 22k Snow ... LRFD)
Now let's solve the problem in ASD.
In ASD,
Fc' = Fc CD CM Ct CP (NDS® Table 5.3.1);
CP = the same equation,
but FcE changes because E min' changes;
E min' = E min' = E min CM Ct (NDS® Table 5.3.1) = 830,000 psi (1.0)(1.0);
E min' = 830,000 psi. (ASD)
The slenderness ratio doesn't change, so
FcE = 0.822 (830,000 psi) / (21.07)2 = 1537 psi. (ASD specific, and way different than the LRFD value)
Fc* = Fc CD CM Ct.
For Snow, CD = 1.15 (NDS® Table 2.3.2), so,
Fc* = 1950 psi (1.15)(1.0)(1.0) = 2243 psi. (LRFD specific)
FcE/Fc*= 1537 psi / 2243 psi = 0.685. (LRFD)
And c = 0.9, still.
CP = (1 + 0.685)/2(0.9) - √{[(1 + 0.685)/2(0.9)]2 - 0.685/0.9} = 0.597. (ASD)
But, whoa, close to the LRFD value (at least in this example).
Fc' = Fc* CP = 2243 psi (0.597) = 1338 psi. (ASD)
The corresponding capacity is,
Fc' A = 1338 psi (30.75 in.2) = 41,100 lb. (ASD)
P total service load = P Dead + P Snow. (ASD)
P total service load = 41,100 lb. (ASD)
(Approx 21k Dead and 21k Snow ... ASD)
In this case the capacity of the column calculates out about the same for both the LRFD and ASD methods (though a lot of the numbers along the way are different).
References
National Design Specification® for Wood Construction, and Supplement Design Values for Wood Construction, 2005, American Forrest and Paper Association, Washington, D.C., www.awc.org.
Minimum Design Loads for Buildings and Other Structures, ASCE 7 (7-05), American Society of Civil Engineers, www.asce.org.
International Building Code, International Code Council, www.iccsafe.org.
Published by Jeff Filler
Consulting Engineer, Educator, Aspiring Writer and Photographer, Husband, Father, and Serious Hunter. View profile
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