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

Example

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.

How StructSuite solves this

StructSuite's free beam design calculator: Floor Joist, Design per NDS. 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).

Steps

1. Step 1: Select member type

   Design consideration: 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).

   In StructSuite: In the "Select member type" section at the top, click the Floor Joist card.

2. Step 2: Select design method

   Design consideration: 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.

   In StructSuite: In the "Select design method" section, click the "Design per 2024 NDS (National Design Specification for Wood Construction)" card.

3. Step 1: Geometry & Configuration

   Design consideration: 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.

   In StructSuite: Open the Geometry section. In the beam schematic, enter span lengths: 14. For each support, use the support type dropdown to select: Pinned, Pinned. NDS 2024 Eq 3.3-1, 3.4-2 for flexure and shear.

4. Step 2: Load Definition (ASCE 7-22 / NDS 2024)

   Design consideration: 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% higher moment capacity—extends max span ~7%.

   In StructSuite: Open the Loads section. Add load items with these exact values: D 12 plf; L 53 plf. For each uniform load: set Type (D, L, Lr, S), enter the plf value in Uniform (plf). For point loads: set category Point, enter magnitude (lb) and position (ft). Include self-weight. NDS 2024 Table 2.3.2 sets CD.

5. Step 3: Material & Section

   Design consideration: Cr=1.15 from NDS §4.3.9. CF from Table 4A for 2×10. Deflection L/360. Must have load distribution (sheathing)—open joists without subfloor don't get Cr. Blocking helps; not required for Cr.

   In StructSuite: Open the Materials section. In Product category select the type; in Species select Douglas Fir-Larch; in Grade select No. 2; in Size select 2×10. Open Design Parameters: set Wet service (CM) to No; set Repetitive member to Yes (Cr = 1.15); select Deflection limit. NDS 2024 §4.3.

Live design (pre-filled)

The form below is the real StructSuite module with example data loaded. Display only—values cannot be changed.