Problem 6.15

 

Given

 

Load	wD = 200 lb/ft wS = 300 lb/ft
Load 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.
Normal temperatures
Bending about strong axis

 

 

Size  (NDS Supplement table 1D)

 

A = 96.25 in.²
I_x = 2,972 in.⁴
S_x = 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
F_vx = 210 psi
E_x = 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 = 0.87 for Fb, 0.8 for Ft, 0.875 for Fv, 0.53 for Fc⊥, 0.73 for Fc, 0.833 for E and Emin 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-bx	2,400 1,450	1.15	0.87	1	1	0.9802	1	1			2,353 --
tension stress parallel to grain	Ft	875	1.15	0.8	1							--
shear stress parallel to grain	Fvx	210	1.15	0.875	1							212
compression stress perpendicular to grain	Fc⊥x	740 650		0.53	1						1	--
compression stress parallel to grain	Fc	1,000	1.15	0.73	1					1		--
radial tension	Frt	--	1.15	--	1							--
modulus of elasticity (or MOE)	Ex	1,700,000		0.833	1							1,416,000
modulus of elasticity for stability calculations	Ex min	880,000		0.833	1							--
modulus of elasticity (or MOE)	Ey	1,300,000		0.833	1							1,083,000
modulus of elasticity for stability calculations	Ey min	670,000		0.833	1							--

 

 

Actual Stresses (ASD) and Deflection

 

vertical ground reactions:  wL/2
V_max = +/- wL/2
M_max = wL²/8

f_b = M/S = (500 lb/ft x 1 ft/12 in.)(20 ft x 12 in/ft)² / (8)(308.8) = 971.5 psi

f_v = 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

Δ_S = 5(300 lb/ft x 1 ft/12 in.)(20 ft x 12 in./ft)⁴ / 384(1,416,000 psi)(2972 in.⁴) = 0.2566 in.

Δ_D+S = 5((200+300 lb/ft) x 1 ft/12 in.)(20 ft x 12 in./ft)⁴ / 384(1,416,000 psi)(2972 in.⁴) = 0.4277 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:  2,353 psi > 971.5 psi  ✓

shear:  212 psi > 77.92 psi  ✓

deflection_S:  0.6667 in. > 0.2566 in.  ✓

deflection_D+S:  1.000 in. > 0.4277 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)