Sawn Timber Column Design: NDS 2018 No.2 SPF Compression Worked Example

Timber columns carry axial gravity loads from beams, girders, and trusses to the foundation. Unlike steel columns where the critical buckling mode is typically global flexural buckling, timber columns are governed by the interaction of material variability (knots, slope of grain, density) with geometric slenderness. The NDS 2018 column stability factor Cp captures this interaction in a single parameter derived from the column's slenderness ratio and the reference compression design value.

This worked example demonstrates a complete compression design check per NDS 2018 Section 3.7 for a 6×6 No.2 Spruce-Pine-Fir (SPF) column carrying an axial load from a roof beam. The column is 12 feet tall with pinned end conditions. You will compute the adjusted compression design value F'c, the column stability factor Cp using the Ylinen buckling formula, and verify the column capacity under both dead and snow load combinations.

PRELIMINARY — NOT FOR CONSTRUCTION. This example is for educational use. All results must be independently verified by a licensed Professional Engineer before use in any design.

Design parameters

Step 1: Reference design values — No.2 SPF

Per NDS 2018 Table 4A (Reference Design Values for Visually Graded Dimension Lumber — 2" to 4" thick), for No.2 SPF in the 6×6 size (posts and timbers, 5×5 and larger, use the same grade values per NDS Supplement Table 4D):

No.2 SPF is the workhorse framing lumber grade in North America for stud walls, posts, and light columns. The reference compression parallel-to-grain value of 1,150 psi reflects the allowable stress for a short column (no buckling reduction). For a 12-ft column, slenderness effects will reduce this substantially.

Step 2: Section properties and loads

Section properties (actual dressed dimensions)

Standard lumber sizes: a nominal 6×6 has actual dressed dimensions of 5.5 × 5.5 inches per NDS Supplement Table 1B (S4S — surfaced four sides).

b = 5.5 in, d = 5.5 in
A = 5.5 × 5.5 = 30.25 in²
Ix = Iy = b × d³ / 12 = 5.5 × 5.5³ / 12 = 5.5 × 166.375 / 12 = 76.2 in⁴
rx = ry = √(I/A) = √(76.2 / 30.25) = √2.519 = 1.587 in

Loads (LRFD)

Per ASCE 7-22, the controlling LRFD combination for dead plus snow on a roof column:

Pu = 1.2 D + 1.6 S = 1.2 × 8,000 + 1.6 × 12,000 = 9,600 + 19,200 = 28,800 lb

Step 3: Adjusted compression design value parallel to grain

Per NDS 2018 Section 3.7.1, the adjusted compression design value is:

F'c = Fc × CD × CM × Ct × CF × Ci × CP

Load duration factor CD

Per NDS Table 2.3.2, the snow load duration factor is CD = 1.15. This applies because the controlling load combination includes snow (duration = 1.15) and the NDS permits CD to be applied to the combined load effect for snow-dominant combinations.

Wet service factor CM

Per NDS Table 4A footnote, No.2 SPF in dry service (MC ≤ 19%) with Fc = 1,150 psi uses CM = 1.0. For wet service, CM = 0.8 for compression parallel to grain for SPF.

Temperature factor Ct

Per NDS Section 3.7.1, Ct = 1.0 for sustained temperatures ≤ 100°F. Interior, conditioned space: Ct = 1.0.

Size factor CF

Per NDS Table 4A footnote, the size factor for compression parallel to grain for a 6×6 (actual 5.5 × 5.5 in) post is CF = 1.0. The size factor for sawn lumber compression members only applies to members thicker than nominally 4 in when both depth and width exceed 4 in; for 6×6, CF = 1.0 per NDS Supplement Tables 4A and 4B.

Incising factor Ci

Per NDS Table 4A, No.2 SPF is generally not incised for interior framing. Ci = 1.0.

Column stability factor CP

Per NDS Section 3.7.1.5, the column stability factor is:

CP = (1 + (FcE/F*c)) / (2c) - √{[(1 + (FcE/F*c)) / (2c)]² - (FcE/F*c) / c}

Where:

• F*c = Fc × CD × CM × Ct × CF × Ci (all factors except Cp)
• FcE = 0.822 × Emin' / (le/d)²
• c = 0.8 for sawn lumber (NDS Section 3.7.1.5)

Compute the intermediate F*c:

F*c = 1,150 × 1.15 × 1.0 × 1.0 × 1.0 × 1.0 = 1,322.5 psi

Compute the effective column length:

Both axes have the same unbraced length and radius of gyration, so either axis governs. Using the strong axis (or weak — identical for a square column):

le = Ke × L = 1.0 × 144 = 144 in
le/d = 144 / 5.5 = 26.18  (NDS uses le/d for rectangular sections; d = least dimension = 5.5 in)

Compute the adjusted modulus of elasticity for stability:

Emin' = Emin × CM × Ct × Ci × CT
Emin' = 510,000 × 1.0 × 1.0 × 1.0 × 1.0 = 510,000 psi

The buckling stiffness factor CT is 1.0 for members without a built-up section or mechanically laminated plies per NDS Section 4.3.1.

Critical buckling design value FcE:

FcE = 0.822 × Emin' / (le/d)² = 0.822 × 510,000 / (26.18)²
FcE = 419,220 / 685.4 = 611.6 psi

Column stability factor CP:

FcE / F*c = 611.6 / 1,322.5 = 0.4625

CP = (1 + 0.4625) / (2 × 0.8) - √{[(1 + 0.4625) / (1.6)]² - 0.4625 / 0.8}
CP = 1.4625 / 1.6 - √{[0.9141]² - 0.5781}
CP = 0.9141 - √{0.8355 - 0.5781}
CP = 0.9141 - √{0.2574}
CP = 0.9141 - 0.5074 = 0.4067

The column stability factor Cp = 0.407, which means the 12-ft column develops only 40.7% of the short-column compression capacity. This is typical for No.2 SPF at an le/d ratio of 26.

Final adjusted compression design value:

F'c = 1,322.5 × 0.4067 = 537.9 psi

Step 4: Compression capacity check

Per NDS 2018, the LRFD factored compression resistance is:

φc = 0.90 (LRFD, NDS Table N3.1, sawn lumber compression parallel to grain)
Pn = F'c × A = 537.9 × 30.25 = 16,271 lb = 16.3 kips
φc Pn = 0.90 × 16.3 = 14.6 kips

Check: Pu = 28,800 lb = 28.8 kips ≥ 14.6 kips → FAIL (utilization = 1.97)

The column is overstressed. The demand-to-capacity ratio of 1.97 means the 6×6 No.2 SPF column at 12 ft is inadequate for this load.

Step 5: Redesign — try 8×8 No.2 SPF

To resolve the overstress, upgrade to an 8×8 (actual 7.25 × 7.25 in) or reduce the unbraced length by adding intermediate lateral bracing. Let's explore the 8×8 option:

Section properties (8×8 actual):

b = 7.25 in, d = 7.25 in
A = 7.25² = 52.56 in²
r = 7.25 / √12 = 7.25 / 3.464 = 2.093 in

For a square section, d in the le/d formula is the least dimension:

le/d = 144 / 7.25 = 19.86
FcE = 0.822 × 510,000 / (19.86)² = 419,220 / 394.4 = 1,063 psi
FcE / F*c = 1,063 / 1,322.5 = 0.8038

CP = (1 + 0.8038) / (2 × 0.8) - √{[(1 + 0.8038) / 1.6]² - 0.8038 / 0.8}
CP = 1.8038 / 1.6 - √{[1.127]² - 1.0048}
CP = 1.1274 - √{1.271 - 1.0048}
CP = 1.1274 - √{0.2662}
CP = 1.1274 - 0.5160 = 0.611

F'c = 1,322.5 × 0.611 = 808 psi
Pn = 808 × 52.56 = 42,468 lb = 42.5 kips
φc Pn = 0.90 × 42.5 = 38.2 kips

Check: 28.8 kips ≤ 38.2 kips → OK (utilization = 0.754)

The 8×8 No.2 SPF column works with a comfortable margin. Note the dramatic improvement from Cp = 0.407 (6×6) to Cp = 0.611 (8×8) — the slenderness ratio improvement from 26.18 to 19.86 accounts for a 50% increase in the stability factor.

Step 6: Alternative — add intermediate bracing to the 6×6

If the 6×6 must be maintained (e.g., for architectural reasons), adding mid-height lateral bracing in both directions halves the effective length:

le = 72 in (mid-height bracing)
le/d = 72 / 5.5 = 13.09

FcE = 0.822 × 510,000 / (13.09)² = 419,220 / 171.4 = 2,446 psi
FcE / F*c = 2,446 / 1,322.5 = 1.849

CP = (1 + 1.849) / 1.6 - √{[(1 + 1.849) / 1.6]² - 1.849 / 0.8}
CP = 2.849 / 1.6 - √{[1.7806]² - 2.311}
CP = 1.7806 - √{3.171 - 2.311}
CP = 1.7806 - √{0.860}
CP = 1.7806 - 0.9274 = 0.853

F'c = 1,322.5 × 0.853 = 1,128 psi
φc Pn = 0.90 × 1,128 × 30.25 / 1,000 = 0.90 × 34.1 = 30.7 kips

Check: 28.8 kips ≤ 30.7 kips → OK (utilization = 0.938)

Mid-height bracing salvages the 6×6 section. The bracing must resist 2% of the column axial load as a lateral force per NDS Section 3.7.1.5 (P_lateral = 0.02 × 28,800 = 576 lb, a modest requirement easily handled by blocking or strapping).

Step 7: Bearing at column base

Per NDS 2018 Section 3.10.4, the compression perpendicular to grain at the column base must be checked. The column bears on a steel base plate:

F'c⟂ = Fc⟂ × CM × Ct × Ci × Cb
F'c⟂ = 425 × 1.0 × 1.0 × 1.0 × 1.0 = 425 psi

For bearing area factor Cb = 1.0 at the end of a member (no additional bearing length beyond the plate):

φc⟂ = 0.90 (LRFD)
Rn = 425 × 52.56 = 22,338 lb = 22.3 kips (> 28.8 kips needed → FAIL)

The bearing perpendicular to grain is inadequate for the bare 8×8 area. A steel bearing plate must distribute the reaction. Required bearing area:

A_req = Pu / (φc⟂ × F'c⟂) = 28,800 / (0.90 × 425) = 28,800 / 382.5 = 75.3 in²

For an 8×8 column, the full end area is only 52.56 in², so a bearing plate larger than the column footprint is required. Use a 10 × 10 × 3/8 in steel plate:

A_plate = 10 × 10 = 100 in² > 75.3 in² → OK

The plate must be thick enough to distribute the load without excessive bending. Per AISC 360:

t_min = √(2.22 × Pu × n² / (Fy_plate × B_plate × N_plate)) = adequate for 3/8 in plate

Results summary

Key takeaways

• The column stability factor Cp dominates timber column design. For the 6×6 at le/d = 26.18, Cp = 0.407, meaning the column develops only 41% of its short-column capacity. Slenderness kills axial capacity nonlinearly — the Ylinen formula in NDS Section 3.7 properly accounts for the nonlinear transition from short-column crushing behavior to Euler buckling.

• Square columns have identical capacity in both axes. For a 6×6 post, rx = ry = 1.587 in. But for rectangular columns (e.g., 4×6), the weak-axis slenderness almost always governs — always check both le/d ratios and apply the larger one to the Cp calculation.

• No.2 grade means knot limitations control. No.2 SPF permits knots up to one-third the face width at the edge of the wide face. For a 6×6, this means knots up to approximately 1.8 inches are permitted. At the knot location, the net section is smaller than the gross section, and combined with the slope-of-grain limits (1:8 for No.2), the effective modulus Emin is reduced relative to Select Structural or No.1 grades.

• Intermediate bracing is more cost-effective than upsizing the member. Adding a single mid-height row of 2×4 blocking between adjacent columns transforms the 6×6 from failing (Cp = 0.407) to working (Cp = 0.853). The blocking cost is negligible compared to upgrading to an 8×8 with the attendant increase in lumber cost, weight, and connection hardware.

• Bearing perpendicular to grain often controls at the column base. Even when the column body has adequate axial capacity, the end-bearing area may be insufficient. The 8×8 solution required a steel bearing plate larger than the column footprint. Always check compression perpendicular at both the top (beam-to-column bearing) and bottom (column-to-foundation bearing) interfaces.

FAQ

What is the difference between No.2 and No.1 SPF for columns?

Per NDS Supplement Table 4A, No.1 SPF has Fc = 1,300 psi (vs 1,150 for No.2) and Emin = 580,000 psi (vs 510,000 for No.2). The 13% higher Fc and 14% higher Emin produce a compound improvement in Cp because both the numerator (Fc) and the FcE term increase. For a 6×6 at 12 ft, No.1 SPF yields FcE/Fc ratio improvement and approximately a 20% higher capacity. However, No.1 SPF is significantly more expensive and less available than No.2 in many markets.

When is a timber column considered "short" (no Cp reduction)?

Per NDS 2018 Commentary C3.7.1.5, a column is effectively "short" (Cp ≥ 0.95) when le/d ≤ 11 for sawn lumber. At le/d = 11, FcE ≈ 3,380 psi and FcE/F*c ≈ 2.56, yielding Cp ≈ 0.95. For a 6×6 (d = 5.5 in), this corresponds to L ≤ 11 × 5.5 / 12 = 5.04 ft. In practice, most timber columns exceed this length, so the Cp reduction is almost always relevant.

How do end conditions affect the effective length?

Per NDS 2018 Appendix G (Column Effective Length), the effective length factor Ke depends on end restraint. For sawn lumber columns: pinned-pinned = 1.0, fixed-fixed = 0.65, fixed-pinned = 0.80, and fixed-free (cantilever) = 2.1. In timber construction, true fixity is difficult to achieve because timber connections are inherently semi-rigid. Most designers conservatively use Ke = 1.0 for pin-ended columns unless the connection restraint can be rigorously justified by testing or advanced analysis.

Can I use glulam columns instead of sawn lumber?

Yes. Per NDS 2018 Table 5A, glulam columns are manufactured with higher reference compression design values — typically Fc = 1,650–2,200 psi for Western Species combinations (vs 1,150 for No.2 SPF sawn). Additionally, glulam columns are available in larger cross-sections (up to 14-1/4 in wide by 60 in deep) and provide both higher capacity and greater architectural flexibility. The design procedure for glulam columns follows NDS Section 5.3.4 and uses the same Cp formulation with c = 0.9 (instead of 0.8 for sawn lumber), which slightly improves the stability factor.