-------------- | ------ | -------------------------------- | -------------------------------- | -------------- | | Load duration | CD | Short-term strength increase | 0.90 (permanent) to 2.0 (impact) | Fb, Ft, Fv, Fr | | Wet service | CM | Moisture content above 19% | 0.67-1.0 (species dependent) | All properties | | Temperature | Ct | Sustained high temperature | 0.50-1.0 (above 100F) | All properties | | Beam stability | CL | Lateral buckling of beams | 0.30-1.0 (depends on Lu/d) | Fb | | Size | CF | Size effect on strength | 0.40-1.50 (smaller is stronger) | Fb, Ft, Fc | | Column stability | CP | Buckling of columns | 0.10-1.0 (depends on le/d) | Fc | | Volume | CV | Volume effect (glulam only) | 0.50-1.0 | Fb (glulam) | | Flat use | Cfu | Flatwise bending | 1.0-1.20 | Fb | | Incising | Ci | Preservative treatment incisions | 0.80-1.0 | All properties | | Repetitive member | Cr | Load sharing in assemblies | 1.0 or 1.15 | Fb | | Bearing area | Cb | Bearing stress distribution | 1.0-1.50 | Fc_perp |

Load duration factor CD values (NDS Table 2.3.2)

The shortest-duration load in the combination controls. CD does NOT apply to modulus of elasticity E or compression perpendicular to grain Fc_perp.

Common Wood Species Design Values

Southern Pine (No. 2 grade, 2x10 at 12% MC)

Douglas Fir-Larch (No. 2 grade, 2x10 at 12% MC)

Glulam 24F-1.8E (balanced layup)

Worked Example — Floor Joist Design

Problem: Select a Southern Pine No. 2 floor joist for a 14 ft span at 16 in on center. Dead load = 12 psf (including joist), live load = 40 psf. Check bending, shear, and deflection.

Step 1 — Applied loads

Tributary width: 16 in = 1.33 ft
w_dead = 12 × 1.33 = 16.0 lb/ft
w_live = 40 × 1.33 = 53.3 lb/ft
w_total = 69.3 lb/ft

CD = 1.00 (10-year live load duration)

Step 2 — Bending check

M = w_total × L² / 8 = 69.3 × 14² / 8 = 1,698 lb-ft = 20,376 lb-in

Required section modulus: S_req = M / (Fb × CD × CF × Cr)
Try 2x10 Southern Pine No. 2:
  Fb = 1,050 psi, CF = 1.0 (for 2x10), Cr = 1.15 (repetitive member)
  Fb' = 1,050 × 1.00 × 1.0 × 1.15 = 1,208 psi

  S = 1.5 × 9.25² / 6 = 21.39 in³ (actual dimensions: 1.5 × 9.25)
  fb = 20,376 / 21.39 = 953 psi < 1,208 psi ✓

  Utilization: 953 / 1,208 = 0.79 → OK

Step 3 — Shear check

V = w_total × L / 2 = 69.3 × 14 / 2 = 485 lb

fv = 3V / (2bd) = 3 × 485 / (2 × 1.5 × 9.25) = 52.5 psi
Fv' = 175 × 1.00 (CD) = 175 psi

fv = 52.5 < 175 ✓ (shear rarely governs for floor joists)

Step 4 — Deflection check

Live load deflection: Δ_LL = 5 × w_live × L⁴ / (384 × E × I)
w_live = 53.3 lb/ft = 4.44 lb/in
I = 1.5 × 9.25³ / 12 = 99.49 in⁴
E = 1,600,000 psi

Δ_LL = 5 × 4.44 × (14×12)⁴ / (384 × 1,600,000 × 99.49)
Δ_LL = 5 × 4.44 × 5.67 × 10⁸ / 6.11 × 10¹⁰ = 0.206 in

L/360 = 168/360 = 0.467 in → 0.206 < 0.467 ✓

Total load deflection:
Δ_total = 5 × (69.3/12) × (168)⁴ / (384 × 1,600,000 × 99.49)
Δ_total = 0.319 in

L/240 = 168/240 = 0.700 in → 0.319 < 0.700 ✓

The 2x10 Southern Pine No. 2 at 16 in OC works for this 14 ft span.

Common Lumber Sizes and Section Properties

Frequently Asked Questions

What is the load duration factor CD in NDS? CD accounts for the fact that wood can sustain higher stresses for short-duration loads. For permanent (dead) loads CD = 0.90, for 10-year occupancy loads CD = 1.00, for 2-month snow CD = 1.15, for 7-day construction CD = 1.25, and for wind/seismic CD = 1.60. The shortest-duration load in the combination determines which CD applies. This is unique to wood design and has no direct equivalent in steel or concrete codes.

What is the difference between sawn lumber and glulam design values? Sawn lumber design values depend on species, grade, and size category (dimension lumber, timbers, decking). Glulam (glued-laminated timber) is manufactured in controlled conditions with graded laminations, so it achieves higher and more reliable design values than equivalent-size sawn timber. Glulam also uses different adjustment factors (volume factor CV instead of size factor CF) and has separate values for positive and negative bending in unbalanced layups.

Why are there so many adjustment factors in NDS? Wood is a natural material whose strength depends on moisture content, temperature, size, duration of load, and manufacturing process. Each adjustment factor accounts for a specific condition that modifies the reference design value established under standard test conditions. While this creates complexity, it also means wood design can be optimized more finely for actual service conditions than a single factor approach would allow.

What is the beam stability factor CL and when does it control? CL accounts for lateral-torsional buckling of beams that are not braced along their compression edge. It applies when the unbraced length of the compression edge exceeds a threshold that depends on the section depth. For a 2x10 with unbraced length exceeding approximately 18 times the depth (about 14 ft), CL drops below 1.0 and reduces the effective bending design value. Most floor joists are braced by sheathing and subfloor, so CL = 1.0.

What is the repetitive member factor Cr? Cr = 1.15 applies when three or more parallel members are spaced at no more than 24 inches on center and are connected by a load-distributing element (sheathing, decking). This factor accounts for load sharing: if one member is weaker than the others, the adjacent members carry more load through the sheathing. It applies only to bending stress Fb. Most floor joist and rafter assemblies qualify for this factor.

How does moisture content affect wood strength? Wood strength decreases as moisture content increases above the fiber saturation point (about 30%). The NDS reference design values are for moisture content of 19% or less (dry conditions). For members that will be exposed to moisture (wet service, MC > 19%), the wet service factor CM reduces the design values by 10-30% depending on the property and species. Southern Pine has relatively high wet-service values compared to other species.

What is the column stability factor CP and how is it calculated? CP is the adjustment factor for columns that accounts for Euler buckling. It reduces the compression design value Fc based on the column slenderness ratio le/d, where le is the effective unbraced length and d is the least dimension of the cross-section. For le/d < 11, CP = 1.0 (short column). For le/d between 11 and 50, CP is computed from the NDS equation that interpolates between the Euler buckling stress and the crushing strength. For le/d > 50, the column is too slender and not permitted by NDS. Most practical wood columns have le/d between 15 and 35.

What is the difference between allowable stress design (ASD) and load and resistance factor design (LRFD) for wood? NDS uses ASD by default, where reference design values are compared directly to service-level (unfactored) stresses. NDS also provides LRFD format in the Appendix, where resistance factors phi and load factors from ASCE 7 are applied. The two methods are calibrated to give similar results, but LRFD provides more uniform reliability across different load ratios. Most wood design in practice uses ASD because it is simpler and the NDS tables are formatted for ASD.

What are engineered wood products and how do they compare to sawn lumber? Engineered wood products include laminated veneer lumber (LVL), parallel strand lumber (PSL), laminated strand lumber (LSL), I-joists, and cross-laminated timber (CLT). These products have higher and more consistent design values than sawn lumber because they are manufactured from graded veneers or strands under controlled conditions. LVL bending stress can reach 2900 psi compared to 1000-1500 psi for sawn lumber. However, engineered products cost 2-4 times more than sawn lumber and require special design considerations (fastener values, bearing lengths, field modifications).

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