Repetitive Member Factor (Cr) for Floor Joists per NDS 2024

Summary: Design floor joists with repetitive member factor Cr = 1.15 per NDS 2024 §4.3.9 when members are 24 in. o.c. or less and load is distributed. Floor…

Overview

Design floor joists with repetitive member factor Cr = 1.15 per NDS 2024 §4.3.9 when members are 24 in. o.c. or less and load is distributed.

Problem statement

Floor joists 2×10 Douglas Fir-Larch No. 2 at 16 in. o.C. Span 14 ft. Dead 12 plf, live 40 psf. Spacing 16 in. < 24 in.—repetitive member Cr = 1.15 applies (NDS 2024 §4.3.9). Check capacity with Cr. CD = 1.0.

Workflow in StructSuite

StructSuite's free beam design calculator: open the Wood Beam module. Span [14], pinned-pinned. Add uniform load—tributary 16/12 ft for 40 psf live. In Design Parameters, check Repetitive member. StructSuite applies Cr = 1.15 to Fb (not Fv, not E). Increases allowable bending stress 15%. Only for dimension lumber 2–4 in. thick, spaced ≤24 in. o.c., with load distribution (sheathing, blocking).

Design considerations (excerpt)

Repetitive members share load through sheathing—adjacent joists help carry concentrated loads. Cr = 1.15 increases Fb 15%; allows longer span or smaller size. Requires: 2–4 in. thick, ≤24 in. o.c., structural sheathing (plywood, OSB). 14 ft at 16 in. o.c.: Cr applies. With Cr, 2×10 may span 14 ft where without Cr would need 2×12. 24 in. o.c. still gets Cr but higher plf load. Spacing > 24 in.: no Cr. Cr applies only to Fb—flexure. No effect on shear, compression, or deflection. Flexure-controlled designs (long spans) benefit most. Single beam, headers, girders: no Cr.

65 plf total (12 D + 53 L). 40 psf live × 1.33 ft = 53 plf. Subfloor + joist ~12 plf. Full span uniform. Cr allows 15%…

Related terms

• StructSuite
• structural engineering
• design example
• wood beam
• NDS 2024
• nds repetitive cr

Open the live example

Use the interactive, read-only wizard: Open this example in StructSuite (pre-filled inputs and step-by-step UI).