Subfloor Load Capacity Calculator
Calculate the allowable uniform load capacity of a subfloor system based on joist dimensions, spacing, span length, and lumber grade properties. Results include total load capacity, dead load allowance, and live load capacity.
Clear distance between supports
On-center spacing between joists (typically 12, 16, or 24 in)
Actual (dressed) width of joist (e.g. 1.5 in for a 2× member)
Actual (dressed) depth of joist (e.g. 9.25 in for a 2×10)
Adjusted allowable bending stress for lumber grade/species (see NDS tables)
Modulus of elasticity for lumber (e.g. 1,600,000 psi for Douglas Fir-Larch No.2)
Estimated dead load (subfloor sheathing + flooring + joists, typically 10–15 psf)
Maximum allowable deflection ratio per IBC/IRC
Formulas Used
Section Modulus: S = b·d² / 6
Moment of Inertia: I = b·d³ / 12
Bending Capacity (uniform load, simple span):
Mallow = F'b × S |
wbend = 8·F'b·S / L² (lb/in per joist)
Converted to psf: wpsf = wlb/in × 12 / sft
Deflection Capacity (uniform load, simple span):
δallow = L / Limit |
wdefl = 384·E·I·δallow / (5·L⁴) (lb/in per joist)
Governing Total Load: wtotal = min(wbend, wdefl)
Live Load Capacity: LL = wtotal − Dead Load
Assumptions & References
- Simple span (single span) beam model — conservative for continuous spans.
- Uniform load distribution across the tributary width of each joist.
- F'b is the adjusted allowable bending stress (already includes all NDS adjustment factors: CD, CM, CF, Cr, etc.).
- Deflection limit applies to live load per IBC Table 1604.3 (L/360 for floors).
- Minimum residential live load: 40 psf per IBC Table 1607.1 / IRC Table R301.5.
- Shear capacity is not checked here — verify separately for short, heavily loaded spans.
- References: NDS 2018 (AWC), IBC 2021, IRC 2021, ASCE 7-22.
- Always verify results with a licensed structural engineer for construction use.