Problem 6.14
Given
Load wD = 200 lb/ft
wS = 300 lb/ftLoad combination D+S (ASD)
1.2D+1.6S (LRFD)Span L = 20 ft Member size 5 x 19-1/4 Glulam bending stress class 24F-1.7E SP Unbraced length lu = 0 Moisture content MC < 16% Live load deflection limit Allow. ΔS ≤ L/360 = 0.6667 in.
Allow. ΔD+S ≤ L/240 = 1.000 in.Dry-service conditions Normal temperatures Bending about strong axis
Size (NDS Supplement table 1D)
A = 96.25 in.2
Ix = 2,972 in.4
Sx = 308.8 in.3
Reference Design Values (most conservative of 24F SP/SP beams in NDS Supplement table 5A Expanded)
F+bx = 2,400 psi
Fvx = 210 psi
Ex = 1,700,000 psi
Ey = 1,300,000 psi
Adjusted ASD Values: F'b, F'v , E'x , E'y
NDS Supplement table 5A CV = (21/L)1/x (12/d)1/x (5.125/b)1/x = (21/20)1/20 (12/19.25)1/20 (5.125/5)1/20 = 0.9802 ≤ 1.0
Cfu = 1
CM = 1
NDS section 2.3
CD = 1.15 (S)
Ct = 1
NDS section 3.3 CL = 1 (lu = 0)
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-bx2,400
1,4501.15 1 1 1 0.9802 1 1 2,705
--tension stress parallel to grain Ft 875 1.15 1 1 -- shear stress parallel to grain Fvx 210 1.15 1 1 242 compression stress perpendicular to grain Fc⊥x 740
6501 1 1 -- compression stress parallel to grain Fc 1,000 1.15 1 1 1 -- radial tension Frt -- 1.15 1 1 -- modulus of elasticity (or MOE) Ex 1,700,000 1 1 1,700,000 modulus of elasticity for stability calculations Ex min 880,000 1 1 -- modulus of elasticity (or MOE) Ey 1,300,000 1 1 1,300,000 modulus of elasticity for stability calculations Ey min 670,000 1 1 --
Actual Stresses (ASD) and Deflection
vertical ground reactions: wL/2
Vmax = +/- wL/2
Mmax = wL2/8fb = M/S = (500 lb/ft x 1 ft/12 in.)(20 ft x 12 in/ft)2 / (8)(308.8) = 971.5 psi
fv = 1.5V/A = 1.5 (500 lb/ft x 1 ft/12 in.)(20 ft x 12 in/ft) / 2(96.25 in.2) = 77.92 psi
ΔS = 5(300 lb/ft x 1 ft/12 in.)(20 ft x 12 in./ft)4 / 384(1,700,000 psi)(2972 in.4) = 0.2138 in.
ΔD+S = 5((200/2+300 lb/ft) x 1 ft/12 in.)(20 ft x 12 in./ft)4 / 384(1,700,000 psi)(2972 in.4) = 0.2850 in. (dead load may be reduce by 1/2 if MC < 16% when installed and kept dry, IBC table 1604.3 footnote d)
Is Member Adequate? (ASD)
bending: 2,705 psi > 971.5 psi ✓
shear: 242 psi > 77.92 psi ✓
deflectionS: 0.6667 in. > 0.2138 in. ✓
deflectionD+S: 1.000 in. > 0.2850 in. ✓
Nominal LRFD Values
Fbn, Fvn, E
Adjusted LRFD Values
F'bn, F'vn, E'
Adjusted LRFD Moment and Shear Resistances
M'n, V'n
Factored Moment and Shear (LRFD) and Actual Deflection
Mu, Vn, Δ
Is Member Adequate? (LRFD)