Problem 6.5

 

Given

 

Load PD = 400 lb
PLr = 1,600 lb
Load combination D+Lr (ASD)
1.2D+1.6Lr (LRFD)
Span L = 8 ft
Member size 4x8
Stress grade and species No. 1 DF-L
Unbraced length lu = L = 8 ft
Moisture content MC ≤ 19%
Live load deflection limit Allow. Δ ≤ L/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 7.25"
A = 25.38 in.2
Sxx = 30.66 in.3
Ixx = 111.1 in.4

 

 

Reference Design Values  (NDS Supplement table 4A)

 

Fb = 1,000 psi
Fv = 180 psi
E = 1,700,000 psi

 

 

Adjusted ASD Values: F'b, F'v , E' 

 

NDS Supplement table 4A

Cr = 1
CM = 1
CF = 1.3 for Fb, 1.2 for Ft, 1.05 for Fc
Cfu = 1

NDS Supplement section 2.3

CD = 1.25 (Lr)
Ct = 1

 NDS Supplement section 3.3

lu/d = 13.24 ≥ 7
le = 1.37 lu + 3d = 153.27 in.
RB = sqrt(led/b2) = sqrt(153.27" x 7.25" / (3.5")2) = 9.524
 F*b = 1,625 psi
FbE = 1.20 E'min / RB2 = 8,202 psi
 CL = 0.9880

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 1,000 1.25 1 1  0.9880 1.3 1 1   1     1,606
tension stress parallel to grain Ft 675 1.25 1 1   1.2       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 1,500 1.25 1 1   1.05     1 1     --
modulus of elasticity (or MOE) E 1,700,000   1 1           1     1,700,000
modulus of elasticity for stability calculations Emin 620,000   1 1           1 1   620,000

 

 

Actual Stresses (ASD) and Deflection

 

fb = M/S = (4 kip-ft x 1000 lb/kip x 12 in/ft) / 30.66 in.3 = 1,566 psi

fv = 1.5V/A = 1.5 (1000 lb) / 25.38 in.2 = 59.10 psi

ΔL = (1600 lb)(96 in.)3 / 48(1,700,000 psi)(111.1 in.4) = 0.1561 in.

 

 

Is Member Adequate? (ASD)

 

bending:  1,606 psi > 1,566 psi  ✓

shear:  225 psi > 59.10 psi  ✓

deflection:  0.2667 in. > 0.1561 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)