Problem 6.24

 

Given

 

closely spaced floor beams

Load	D = 18 psf L = 50 psf
Load combination	D+L (ASD) 1.2D+1.6L (LRFD)
Span	L = 14 ft
Member spacing	Tributary width = b = 16 in. OC wD = 18 lb/ft2 (1 ft/12 in.)2 x 16 in. = 2 lb/in. wL = 50 psf (1 ft/12 in.)2 x 16 in. = 5.556 lb/in.
Stress grade and species	No. 1 Hem-Fir
Unbraced length	lu = 0
Moisture content	MC ≤ 19%
Live load deflection limit	Allow. ΔL ≤ L/360 = 0.4667 in. Allow. ΔD+L ≤ L/240 = 0.7000 in.
Dry-service conditions
Normal temperatures
Bending about strong axis

 

 

Trial Size

 

F_b = 975 psi  (NDS Supplement table 4A)

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

S ≥ M_max/F_b = (7.556 lb/in.) x (14 ft x 12 in./ft)² / (8 x 975 lb/in.²) = 27.34 in.³

S_2x10 = 21.39 in.³  ✘  (NDS Supplement table 1B)
S_2x12 = 31.64 in.³  ✓  (NDS Supplement table 1B)

 

 

Size Category  (NDS Supplement tables 1A and 1B)

 

dimension lumber
dressed size: 1.5" x 11.25"
A = 16.88 in.²
S_xx = 31.64 in.³
I_xx = 178.0 in.⁴

 

 

Reference Design Values  (NDS Supplement table 4A)

 

F_b = 975 psi
F_v = 150 psi
E = 1,500,000 psi

 

 

Adjusted ASD Values: F'_b, F'_v , E' 

 

NDS Supplement table 4A Cr = 1.15 CM = 1 Cfu = 1 CF = 1 NDS Supplement section 2.3 CD = 1 (L) Ct = 1

NDS Supplement section 3.3 CL = 1 (lu = 0) NDS Supplement section 3.7 CP = 1 (not a column) NDS Supplement section 3.10 Cb = 1 (not enough info given)

NDS Supplement section 4.3 Ci = 1 NDS Supplement section 4.4 CT = 1 (not a truss)

 

Property		Reference Design Values (psi) (Table 4A)	Adjustment Factors (Table 4.3.1)											Adjusted Design Values (psi)
			CD	CM	Ct	CL	CF	Cfu	Cr	CP	Ci	CT	Cb
bending stress	Fb	975	1	1	1	1	1	1	1.15		1			1,121
tension stress parallel to grain	Ft	625	1	1	1		1				1			--
shear stress parallel to grain	Fv	150	1	1	1						1			150
compression stress perpendicular to grain	Fc⊥	405		1	1						1		1	--
compression stress parallel to grain	Fc	1,350	1	1	1		1			1	1			--
modulus of elasticity (or MOE)	E	1,500,000		1	1						1			1,500,000
modulus of elasticity for stability calculations	Emin	550,000		1	1						1	1		--

 

 

Actual Stresses (ASD) and Deflection

 

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

f_b = M/S = (7.556 lb/in.) x (14 ft x 12 in./ft)² / (8 x 31.64 in.³) = 842.5 psi

f_v = 1.5V/A = 1.5 (7.556 lb/in. x 14 ft x 12 in./ft) / (2 x 16.88 in.²) = 56.40 psi

Reduced Shear  (NDS section 3.4.3)

f_{v reduced} = 1.5 (7.556 lb/in. x (14 ft x 12 in./ft - 2 x 11.25 in.)) / (2 x 16.88 in.²) = 48.85 psi

See formula 11 for the deflection.

Δ_L = 5(5.556 lb/in.)(14 ft x 12 in./ft)⁴ / 384(1,500,000 psi)(178.0 in.⁴) = 0.2158 in.

Δ_D+L = 5(2/2+5.556 lb/ft)(14 ft x 12 in./ft)⁴ / 384(1,500,000 psi)(178.0 in.⁴) = 0.2547 in.  (dead load may be reduce by 1/2 if MC < 16% when installed and kept dry, IBC table 1604.3 footnote d)

 Δ_D+L = 5(2+5.556 lb/ft)(14 ft x 12 in./ft)⁴ / 384(1,500,000 psi)(178.0 in.⁴) = 0.2935 in.  (if 16% < MC < 19%)

 

 

Is Member Adequate? (ASD)

 

bending:  1,121 psi > 842.5 psi  ✓

shear:  150 psi > 56.40 psi  ✓  (ok, so no need to consider reduced shear)

deflection_L:  0.4667 in. > 0.2158 in.  ✓

deflection_D+L:  0.7000 in. > 0.2547 or 0.2935 in.  ✓

Notches

NDS section 3.2.3, NDS section 5.4.4

At the ends of a glulam, notches on the tension side cannot exceed the lesser of 1/10 of the depth or 3 in.

11.25 in. / 10 = 1.125 in. > 1 in.  ✓

At the ends of a glulam, notches on the compression side cannot exceed 2/5 of the depth.

Compression-side notches cannot extend into the middle third of the span.

NDS section 3.4.3

V_max = wL/2 = 7.556 lb/in. x 14 ft x 12 in./ft / 2 = 634.7 lb

V '_r = (2F'_vbd_n/3)(d_n/d)² = (2 x 150 psi x 1.5 in. x (11.25 -1 in.)/3) ((11.25-1 in.)/11.25 in.)² = 1276 lb

1276 lb > 634.7 lb  ✓

 

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)