OpenSpecEngineering

Checks the bending and shear stresses in a rectangular sawn-lumber beam, joist or header against the adjusted design values of the NDS, by allowable stress design (ASD) [1]. Pick a nominal size and a species and grade, and the page starts from the reference values for visually graded dimension lumber in NDS Supplement Table 4A [3] and applies the adjustment factors. Or choose Custom and enter your own section and adjusted values. Designers and plan checkers use it to confirm or size a member.

The equations

f_b = M/S,   S = bd²/6
f_v = 1.5V/(bd)

M is the applied moment and V the applied shear, both from service (ASD) loads. b is the beam’s width and d its depth, the dressed dry dimensions of NDS Supplement Table 1B [3] when you pick a nominal size. S is the section modulus. The 1.5 is the peak shear stress at the neutral axis of a rectangle.

The adjusted design values come from NDS Table 4.3.1 [1]:

F_b′ = F_b·C_D·C_M·C_F·C_r
F_v′ = F_v·C_D·C_M

C_D is the load duration factor: 1.0 for occupancy live load, 1.15 for snow, 1.6 for wind. C_M is the wet service factor, applied when the wood stays above 19 % moisture. C_F is the size factor for the nominal depth. C_r = 1.15 applies to three or more members at 24 in on center or less that share the load, such as floor joists. The beam passes when both ratios f/F′ are 1 or less. The page also gives E′ for a deflection check.

Limits and common mistakes

• The beam stability factor C_L is taken as 1, which needs the compression edge braced, for example by sheathing nailed to the top of the joists. The page also takes the temperature factor C_t and the incising factor C_i as 1 and assumes the load bears on the narrow face.
• Pick the load duration of the shortest load in the combination; dead load alone uses 0.9 [1].
• Table 4A covers dimension lumber 2 to 4 in thick. For timbers, glulam and engineered lumber, use Custom.
• Shear may be taken at a distance d from the support face [1], so the full end shear is conservative.

Worked example

Floor joists of 2×10 Douglas Fir-Larch No. 2 at 16 in on center span 4.0 m. With 2.4 kPa of total load on a 0.406 m strip, each joist carries 0.975 kN/m, so M = wL²/8 = 1.95 kN·m and V = wL/2 = 1.95 kN.

1. Enter Applied Moment 1.95 and Applied Shear 1.95.
2. Choose Nominal Size 2×10 and Species and Grade Douglas Fir-Larch, No. 2.
3. Choose Load Duration Ten years, Service Condition Dry and Repetitive Member Yes.
4. Press Solve. The Solution shows: Bending stress fb: 5.56301 MPa
   Bending ratio fb / Fb': 0.708693
   Shear stress fv = 1.5V/(bd): 0.326757 MPa
   Shear ratio fv / Fv': 0.26329
   Result: OK
   Adjusted Fb' = Fb·CD·CM·CF·Cr: 7.84968 MPa

Check in US units: S = 1.5 × 9.25² / 6 = 21.39 in³, and 1.95 kN·m is 17,259 lb·in, so f_b = 807 psi. F_b′ = 900 × 1.1 × 1.15 = 1,138.5 psi, and 807/1,138.5 = 0.709.