Problem 5.11
5-1/8 x 28.5 24F-1.8E Douglas Fir, 32' span, D+S, uniform load over simple span, supported against buckling
Cross Section
centroid = 14.25 in.
Ix = bh3/12 = 5.125 x 28.53 / 12 = 9887 in.4 ✓
Sx = I / centroid = 693.8 in.3 ✓
Adjusted Design Values
NDS Supplement table 5A
CV = (21/L)1/x (12/d)1/x (5.125/b)1/x = (21/32)1/10 (12/28.5)1/10 (5.125/5.125)1/10 = 0.8793 ≤ 1.0
Cfu = 1
CM = 1
NDS section 2.3
CD = 1.15 (S)
Ct = 1
NDS section 3.3
CL = 1 (supported against buckling)
NDS section 3.7
CP = 1 (not a column)
NDS section 3.10.4
Cb = 1 (not enough info)
NDS section 5.3
Cc = 1 (not curved)
Property Reference Design
Values (psi)
(Table 5A)Adjustment Factors (Table 5.3.1) Adjusted Design
Values (psi)CD CM Ct CL CV Cfu Cc CP Cb bending stress F+bx
F-bx2400
14501.15 1 1 1 0.8793 1 1 2427
1466tension stress parallel to grain Ft 1100 1.15 1 1 1265 shear stress parallel to grain Fvx 265 1.15 1 1 305 compression stress perpendicular to grain Fc⊥x 650 1 1 1 650 compression stress parallel to grain Fc 1600 1.15 1 1 1 1840 radial tension Frt -- 1.15 1 1 -- modulus of elasticity (or MOE) Ex 1,800,000 1 1 1,800,000 modulus of elasticity for stability calculations Ex min 930,000 1 1 930,000
Adjusted Design Values for MC > 16%
NDS Supplement table 5A
CM = 0.8 for Fb and Ft, 0.875 for Fv, 0.53 for Fc⊥, 0.73 for Fc, 0.833 for E and Emin
Property Reference Design
Values (psi)
(Table 5A)Adjustment Factors (Table 5.3.1) Adjusted Design
Values (psi)CD CM Ct CL CV Cfu Cc CP Cb bending stress F+bx
F-bx2400
14501.15 0.8 1 1 0.8793 1 1 1941
1173tension stress parallel to grain Ft 1100 1.15 0.8 1 1012 shear stress parallel to grain Fvx 265 1.15 0.875 1 267 compression stress perpendicular to grain Fc⊥x 650 0.53 1 1 345 compression stress parallel to grain Fc 1600 1.15 0.73 1 1 1343 radial tension Frt -- 1.15 1 1 -- modulus of elasticity (or MOE) Ex 1,800,000 0.833 1 1,500,000 modulus of elasticity for stability calculations Ex min 930,000 0.833 1 775,000