Problem 6.26

 

Given

 

roof rafters at 6 ft OC
D = 12 psf
L_r = 20 psf  (ASCE 7 table 4-1 or IBC table 1607.1)
No. 2 DF-L
fully braced
C_M = 1
C_t = 1

 

 

Loads

 

horizontal span = 14 ft
height = 4/12 x 14 ft = 4.667 ft  (similar triangles)
diagonal length = sqrt(14² + 4.667²) = 14.76 ft
D' = 14.76D/14 = 12.65 psf
w_D = (12.65 lb/ft²)(1 ft/12 in.)²(6 ft x 12 in./ft) = 6.325 lb/in.

A_T = 14 ft x (6 ft) = 84 ft² < 200 ft²  (ASCE 7 section 4.9.1 or IBC section 1607.11)
R₁ = 1
R₂ = 1  (F ≤ 4)
L_r = 1 x 1 x 20 psf = 20 psf
w_Lr = (20 lb/ft²)(1 ft/12 in.)²(6 ft x 12 in./ft) = 10 lb/in.

 

 

Trial Size

 

F_b = 900 psi

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

S ≥ M_max/F_b = (16.325 lb/in.) x (14 ft x 12 in./ft)² / (8 x 900 lb/in.²) = 63.99 in.³

S_2x14 = 43.89 in.³  ✘  (NDS Supplement table 1B)
S_4x10 = 49.91 in.³  ✘  (NDS Supplement table 1B)
S_4x12 = 73.83 in.³  ✓  (NDS Supplement table 1B)

Let's see if we can use something smaller than 4x12 by using F'_b instead of F_b in our estimate.  The solutions manual uses a 2x16, but that is not in NDS table 1B, so we will try something else.

 

 

Adjusted ASD Values

 

NDS Supplement table 4A Cr = 1 CM = 1 Cfu = 1 CF 2x14 = 0.9 for Fb, 0.9 for Ft, 0.9 for Fc CF 4x10 = 1.2 for Fb, 1.1 for Ft, 1 for Fc CF 4x12 = 1.1 for Fb, 1 for Ft, 1 for Fc NDS Supplement section 2.3 CD = 1.25 (Lr) Ct = 1

NDS Supplement section 3.3 CL = 1 (fully braced) 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	900	1.25	1	1	1	0.9 1.2 1.1	1	1		1			1,013 for 2x14 1,350 for 4x10 1,238 for 4x12
tension stress parallel to grain	Ft	575	1.25	1	1		0.9 1.1 1				1			-- --
shear stress parallel to grain	Fv	180	1.25	1	1						1			225
compression stress perpendicular to grain	Fc⊥	625		1	1						1		1	--
compression stress parallel to grain	Fc	1,350	1.25	1	1		0.9 1 1			1	1			-- --
modulus of elasticity (or MOE)	E	1,600,000		1	1						1			--
modulus of elasticity for stability calculations	Emin	580,000		1	1						1	1		--

 

 

Trial Size

 

F'_{b 2x14} = 1,013 psi
F'_{b 4x10} = 1,350 psi
F'_{b 4x12} = 1,238 psi

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

S_2x14 ≥ M_max/F'_{b 2x8} = (16.325 lb/in.) x (14 ft x 12 in./ft)² / (8 x 1,013 lb/in.²) = 56.86 in.³
S_4x10 ≥ M_max/F'_{b 2x8} = (16.325 lb/in.) x (14 ft x 12 in./ft)² / (8 x 1,350 lb/in.²) = 42.66 in.³
S_4x12 ≥ M_max/F'_{b 2x10} = (16.325 lb/in.) x (14 ft x 12 in./ft)² / (8 x 1,238 lb/in.²) = 46.52 in.³

S_2x14 = 43.89 in.³  ✘  (NDS Supplement table 1B)
S_4x10 = 49.91 in.³  ✓  (NDS Supplement table 1B)
S_4x12 = 73.83 in.³  ✓  (NDS Supplement table 1B)

∴ 4x10 or larger should be okay for bending.

 

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

 

dimension lumber
dressed size: 3.5" x 9.25"
A = 32.38 in.²
S_xx = 49.91 in.³
I_xx = 230.8 in.⁴

 

 

Actual Stresses (ASD) and Deflection

 

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

f_b = M/S = (16.325 lb/in.) x (14 ft x 12 in./ft)² / (8 x 49.91 in.³) = 1,154 psi

f_v = 1.5V/A = 1.5 (16.325 lb/in. x 14 ft x 12 in./ft) / (2 x 32.38 in.²) = 63.53 psi

f_{v reduced} = 1.5 (16.325 lb/in. x (14 ft x 12 in./ft - 2 x 9.25 in.)) / (2 x 32.38 in.²) = 56.53 psi  (NDS section 3.4.3)

 

 

Is Member Adequate? (ASD)

 

bending:  1,350 psi > 1,154 psi  ✓

shear:  225 psi > 63.53 psi  ✓  (ok, so no need to consider reduced shear)

Notch dimensions not given in image, so they cannot be checked.

The 2x16 used in the solutions manual was okay for both bending and shear.

 

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)