Problem 5.15
Same as 5.11 except loaded about minor axis
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 = 2.5625 in.
Ix = bh3/12 = 28.5 x 5.1253 / 12 = 319.7 in.4 ✓
Sx = I / centroid = 124.8 in.3 ✓
Adjusted Design Values
NDS Supplement table 5A
CV = 1 (not loaded ⊥ to wide faces of lams)
Cfu = 1.10
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 Fby 1450 1.15 1 1 1 1 1.10 1 1834 tension stress parallel to grain Ft 1100 1.15 1 1 1265 shear stress parallel to grain Fvy 230 1.15 1 1 265 compression stress perpendicular to grain Fc⊥y 560 1 1 1 560 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) Ey 1,600,000 1 1 1,600,000 modulus of elasticity for stability calculations Ey min 830,000 1 1 830,000
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 Fbx 1450 1.15 0.8 1 1 1 1.10 1 1467 tension stress parallel to grain Ft 1100 1.15 0.8 1 1012 shear stress parallel to grain Fvx 230 1.15 0.875 1 231 compression stress perpendicular to grain Fc⊥x 560 0.53 1 1 297 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 1,600,000 0.833 1 1,330,000 modulus of elasticity for stability calculations Ex min 830,000 0.833 1 690,000