Wood Beam Span Calculator
Verify whether a wood beam is adequate for your project. Enter the span length, select the beam size (width and depth from standard lumber dimensions), choose the wood species and grade, set the loading condition (interior floor, bedroom, exterior, or roof), specify tributary width, deflection limit (L/360, L/240, L/480, L/720), and moisture condition. Results include bending stress, shear stress, and deflection with pass/fail for each, plus utilization ratios, section properties, approximate maximum span, and maximum load capacity. Based on NDS 2018 allowable stress design.
Clear span between supports (center-to-center)
Half the distance to the next beam on each side (floor area supported by this beam)
M_max = wL² / 8 (uniform load)
fb = M / S ≤ Fb′ where S = bd² / 6
V_max = wL / 2
fv = 1.5V / A ≤ Fv′
Δ = 5wL⁴ / (384EI) ≤ L/360
Where:
w = uniform load (plf)
L = span (ft)
b = width (in)
d = depth (in)
E = modulus of elasticity
Fb = allowable bending
Fv = allowable shear
CM = wet service factor (0.85 wet)
Span: 12 ft, Tributary: 8 ft, Interior floor (40+10 psf)
Total load w = 50 psf × 8 = 400 plf
M_max = 400 × 12² / 8 = 7,200 ft-lbs = 86,400 in-lbs
S = 3.5 × 9.25² / 6 = 49.9 in³
fb = 86,400 / 49.9 = 1,731 psi
Fb′ (DF #2) = 900 psi × 1.0 (dry) = 900 psi
Bending: 1,731 > 900 → FAIL ✗
Try 6×12 (5.5×11.25″): S = 116 in³
fb = 86,400 / 116 = 745 psi ✓
Shear: fv = 1.5 × 2,400 / (5.5×11.25) = 58 psi < 180 ✓
Deflection: ~0.24″ < 0.40″ ✓
Result: 6×12 DF #2 PASSES at 12-ft span
How is wood beam span calculated?
Wood beam span is calculated using allowable bending stress (Fb) and allowable shear stress (Fv) for the specific wood species and grade. The maximum moment M = wL²/8 for a uniformly loaded simply supported beam. Section modulus S = bd²/6. Required S = M/Fb. The actual S must be ≥ required S. Deflection is checked using Δ = 5wL⁴/(384EI) and must be within L/360 for floors. Beam size, species, grade, loading, and spacing all affect the maximum allowable span. Always consult a structural engineer for critical applications.
What size beam do I need for a 12-foot span?
For a typical 12-ft span with standard floor loading (40 psf live, 10 psf dead) and 8-ft tributary width: A Douglas Fir #2 beam would need to be at least 4×10 (3.5×9.25") or 6×8 (5.5×7.25"). Two 2×10s sistered together (3×9.25") would work for lighter loads. For a roof with 30 psf snow load: a 4×8 (3.5×7.25") beam may suffice. Span tables from the American Wood Council (AWC) provide quick reference: a 4×10 DF #2 at 12-ft span can support approximately 60 psf total load with a 10-ft tributary width.
What is the difference between SPF and Douglas Fir for beams?
Douglas Fir-Larch (#2) has higher design values than SPF (Spruce-Pine-Fir #2): Fb (bending): DF = 900 psi vs SPF = 875 psi. Fv (shear): DF = 180 psi vs SPF = 135 psi. E (modulus): DF = 1.6M psi vs SPF = 1.4M psi. This means a Douglas Fir beam can span slightly further or carry more load than the same size SPF beam. Southern Yellow Pine #2 is comparable to DF with Fb = 1,200 psi. For a 4×10 beam at 12-ft span, DF allows about 10% more load than SPF. Cost difference is usually minimal — use DF or SYP for critical beams.
What is tributary width and why does it matter?
Tributary width is the width of floor or roof area that transfers load to a specific beam. For beams spaced at 8 ft on center, each beam supports 4 ft on each side = 8 ft tributary width. Total load on the beam = tributary width × span length × load per square foot. A beam with 12-ft tributary width carries 50% more load than one with 8-ft tributary width. Exterior walls have a tributary width of half the floor span. Ridge beams have tributary widths from both roof slopes. Accurate tributary width is essential — underestimating it can lead to undersized beams and structural failure.
🔗 Related Calculators
📐 Formula
M_max = wL² / 8 (uniform load)
fb = M / S ≤ Fb′ where S = bd² / 6
V_max = wL / 2
fv = 1.5V / A ≤ Fv′
Δ = 5wL⁴ / (384EI) ≤ L/360
Where:
w = uniform load (plf)
L = span (ft)
b = width (in)
d = depth (in)
E = modulus of elasticity
Fb = allowable bending
Fv = allowable shear
CM = wet service factor (0.85 wet)
📝 Example Calculation
Span: 12 ft, Tributary: 8 ft, Interior floor (40+10 psf)
Total load w = 50 psf × 8 = 400 plf
M_max = 400 × 12² / 8 = 7,200 ft-lbs = 86,400 in-lbs
S = 3.5 × 9.25² / 6 = 49.9 in³
fb = 86,400 / 49.9 = 1,731 psi
Fb′ (DF #2) = 900 psi × 1.0 (dry) = 900 psi
Bending: 1,731 > 900 → FAIL ✗
Try 6×12 (5.5×11.25″): S = 116 in³
fb = 86,400 / 116 = 745 psi ✓
Shear: fv = 1.5 × 2,400 / (5.5×11.25) = 58 psi < 180 ✓
Deflection: ~0.24″ < 0.40″ ✓
Result: 6×12 DF #2 PASSES at 12-ft span