Problem 6.20

 

Given

 

Load	PD = 400 lb PL = 1600 lb
Load combination	D+L (ASD) 1.2D+1.6Lr (LRFD)
Span	L1 = 8 ft L2 = 4 ft
Member size	4x12
Stress grade and species	Sel. Str. SP
Unbraced length	lateral support is provided at the vertical supports and at the free end lu = 8 ft for main span lu = 4 ft for free end
Moisture content	MC ≤ 19%
Live load deflection limit	Allow. Δfree end ≤ L2/180 = 0.2667 in. Allow. Δbetween supports ≤ L1/360 = 0.2667 in.
Dry-service conditions
Normal temperatures
Bending about strong axis

 

 

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

 

dimension lumber
dressed size: 3.5" x 11.25"
A = 39.38 in.²
S_xx = 73.83 in.³
I_xx = 415.3 in.⁴

 

 

Reference Design Values  (NDS Supplement table 4B)

 

F_b = 2,850 psi
F_v = 175 psi
E = 1,800,000 psi

 

 

Adjusted ASD Values: F'_b, F'_v , E' 

 

NDS Supplement table 4B CF = 1.1 for Fb, 1 for others Cr = 1 Cfu = 1 CM = 1 NDS Supplement section 2.3 CD = 1 (L) Ct = 1

NDS Supplement section 3.3 main span lu/d = 8.533 le = 1.63 lu + 3d = 190.2 in. RB = sqrt(led/b2) = sqrt(190.2" x 11.25" / (3.5")2) = 13.22 F*b = 2,090 psi FbE = 1.20 E'min / RB2 = 4,532 psi CL = 0.9617 free end lu/d = 4.267 le = 2.06 lu = 98.88 in. RB = sqrt(led/b2) = sqrt(98.88" x 11.25" / (3.5")2) = 9.529 F*b = 2,090 psi FbE = 1.20 E'min / RB2 = 8,722 psi CL = 0.9848 0.9617 < 0.9848 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 4B)	Adjustment Factors (Table 4.3.1)											Adjusted Design Values (psi)
			CD	CM	Ct	CL	CF	Cfu	Cr	CP	Ci	CT	Cb
bending stress	Fb	1,900	1	1	1	0.9617	1.1	1	1		1			2,010
tension stress parallel to grain	Ft	1,600	1	1	1		1				1			--
shear stress parallel to grain	Fv	175	1	1	1						1			175
compression stress perpendicular to grain	Fc⊥	565		1	1						1		1	--
compression stress parallel to grain	Fc	2,100	1	1	1		1			1	1			--
modulus of elasticity (or MOE)	E	1,800,000		1	1						1			1,800,000
modulus of elasticity for stability calculations	Emin	660,000		1	1						1	1		660,000

 

 

Actual Stresses (ASD) and Deflection

 

See formula 8 for the deflection of the simply-supported portion of the beam and formulas 7 and 20 for the cantilever portion.

f_b = M/S = (96,000 in-lb) / 73.83 in.³ = 1300 psi

f_v = 1.5V/A = 1.5 (2000 lb) / 39.38 in.² = 76.18 psi

Δ_{main span} = 0.07592 in.

Δ_{free end} = 0.2961 in.

 

 

Is Member Adequate? (ASD)

 

bending:  2,010 psi > 1,300 psi  ✓

shear:  175 psi > 76.18 psi  ✓

deflection in main span:  0.2667 in. > 0.07592 in.  ✓

deflection at free end:  0.2667 in. > 0.2961 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)