Problem 6.18

 

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 lu = 0
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
Sxx = 73.83 in.3
Ixx = 415.3 in.4

 

 

Reference Design Values  (NDS Supplement table 4B)

 

Fb = 2,850 psi
Fv = 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

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 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 1 1.1 1 1   1     2,090
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   --

 

 

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.

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

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

Δmain span = 0.07592 in.

Δfree end = 0.2961 in.

 

 

Is Member Adequate? (ASD)

 

bending:  2,090 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)