Wood Beam Sizing Calculator (Uniform Load)
Size a simply supported wood beam or header under a uniform load: bending moment, required section modulus from allowable bending stress, deflection check against L/360 or L/240, shear check, and the smallest standard sawn or built-up lumber size that passes.
How it works
w = tributary × (LL + DL) + line load; M = wL²/8; S_req = 12M ÷ Fb; Δ = 5wL⁴ ÷ (384EI) ≤ L/limit → I_req; V = wL/2, A_req = 1.5V ÷ Fv; S = bd²/6, I = bd³/12
A beam under a uniform load must satisfy three checks. Bending: the maximum moment at mid-span, wL²/8, divided by the allowable bending stress gives the section modulus the beam needs, and section modulus grows with the square of depth — which is why a deeper beam beats a wider one. Deflection: the live-load sag must stay under the span divided by 360 for floors (240 for roofs and decks), which depends on the moment of inertia and stiffness E and often governs for long spans and low-grade lumber. Shear and bearing rarely govern except for short, heavily loaded headers. Allowable stresses come from the NDS supplement and are adjusted for duration, moisture, size and repetitive use; the defaults here are conservative #2 values. Anything carrying concentrated loads, point loads from posts, or roof snow belongs with an engineer or the manufacturer's software.
Worked example
A 10 ft beam picking up 8 ft of floor at 40 + 15 psf carries 440 plf: moment 5,500 ft·lb, so with Fb = 1,000 psi it needs S ≥ 66 in³; L/360 live-load deflection at E = 1.4×10⁶ needs I ≥ 154 in⁴. A (2) 2×12 (S 63) just misses on bending, a (3) 2×10 (S 64) too, so a (3) 2×12 (S 95, I 534) or a 3½ × 9¼ LVL is the answer.
Frequently asked questions
What Fb should I use?
From the NDS supplement for your species and grade, times the adjustment factors: wet service (0.85 for Fb on decks), load duration (1.15 snow, 1.25 construction), size factor (1.0–1.5 for 2× depths) and repetitive member (1.15 when 3+ members at ≤ 24″). #2 SPF base is 875 psi; #2 southern pine 2×12 about 975; LVL 2,600.
Why does deflection govern more than strength?
Because strength requirements scale with L² but deflection with L⁴ — double the span and the beam needs 4× the section modulus but 16× the stiffness. Bouncy floors are stiffness problems, not strength problems.
Related calculators
Simplified allowable-stress design for a simply supported beam under uniform load. Concentrated loads, continuous spans, lateral stability, notches and connections are not covered; have any structural beam checked by a licensed engineer or sized with the manufacturer's software.