Wood Allowable Bending Stress

F_{b}^{'} = F_{b} C_{D} C_{M} C_{t} C_{L} C_{F} C_{f u} C_{i} C_{r}

NDS allowable stress design: published Fb is multiplied by load-duration, wet-service, temperature, beam-stability, size, flat-use, incising, and repetitive-member factors. Then fb=M/S ≤ Fb'.

$F_{b}$ (psi): Base allowable bending
$C_{D}$: Load-duration factor

Step-by-step solved example

Fb=1000 psi, CD=1.15 (snow), other C=1. S=30 in³, M=2.4 kip·ft. Check.

Wood beam: fb = M/S on a rectangular section.

1. Fb'
1000×1.15=1150 psi.
2. fb
M=2.4×12000=28800 lb·in, fb=28800/30=960 psi < 1150, OK.

Answer: fb=960 psi < Fb'=1150 psi (OK)

10 practice questions

0/10 correct

1.CD for 10-year occupancy live load is

2.CD for impact/wind (short) is larger because

3.Wet-service CM < 1 when

4.Section modulus S=

5.Repetitive-member Cr=1.15 applies to

6.Fv' shear check uses

7.E used in deflections is also adjusted by

8.Size factor CF < 1 for

9.Fb=900, all C=1, S=20 in³: φM? (ASD) allowable M =

10.LRFD wood uses