Wood Beam fb = M/S vs Fb'

f_{b} = \frac{M}{S} \leq F_{b}^{'}, S = \frac{b d ^{2}}{6} (rectangle)

Sawn rectangular lumber: S=bd²/6 with d the depth. Compare with adjusted Fb' (duration, wet, size, repetitive…). Also check Fv', deflection L/360, and stability CL.

$b, d$: Width and depth
$S$: Section modulus

Step-by-step solved example

2×10 S4S (1.5×9.25 in), M=1500 lb·ft, Fb'=1000 psi. OK?

Wood joist: fb = M/S, S = bd²/6.

1. S
1.5×9.25²/6=21.4 in³.
2. fb
M=18000 lb·in, fb=18000/21.4=841 psi < 1000, OK.

Answer: fb ≈ 841 psi < 1000 (OK)

10 practice questions

0/10 correct

1.Nominal 2×10 dressed is about

2.If d doubles, S

3.Shear fv=1.5 V/(b d) ≤ Fv'

4.Live-load deflection limit often

5.CL beam-stability factor <1 when

6.M=S Fb' with S=20 in³, Fb'=1200 psi: M=

7.Glulam uses similar form with

8.Repetitive Cr=1.15 is for

9.E for deflection of sawn lumber is also multiplied by

10.Notching at the support reduces