Problem 6.14

 

Given

 

Load wD = 200 lb/ft
wS = 300 lb/ft
Load combination D+S (ASD)
1.2D+1.6S (LRFD)
Span L = 20 ft
Member size 5 x 19-1/4
Glulam bending stress class 24F-1.7E SP
Unbraced length lu = 0
Moisture content MC < 16%
Live load deflection limit Allow. ΔSL/360 = 0.6667 in.
Allow. ΔD+SL/240 = 1.000 in.
Dry-service conditions
Normal temperatures
Bending about strong axis

 

 

Size  (NDS Supplement table 1D)

 

A = 96.25 in.2
Ix = 2,972 in.4
Sx = 308.8 in.3

 

 

Reference Design Values  (most conservative of 24F SP/SP beams in NDS Supplement table 5A Expanded)

 

F+bx = 2,400 psi
Fvx
= 210 psi
Ex
= 1,700,000 psi
Ey = 1,300,000 psi

 

 

Adjusted ASD Values: F'b, F'v , E'x , E'y

 

NDS Supplement table 5A

CV = (21/L)1/x (12/d)1/x (5.125/b)1/x = (21/20)1/20 (12/19.25)1/20 (5.125/5)1/20 = 0.9802 ≤ 1.0

Cfu = 1

CM = 1

NDS section 2.3

CD = 1.15 (S)

Ct = 1

NDS section 3.3

CL = 1 (lu = 0)

NDS section 3.7

CP = 1 (not a column)

NDS section 3.10.4

Cb = 1 (not enough info)

NDS section 5.3

Cc = 1 (not curved)

 

Property Reference Design
Values (psi)
(Table 5A)
Adjustment Factors (Table 5.3.1) Adjusted Design
Values (psi)
CD CM Ct CL CV Cfu Cc CP Cb
bending stress F+bx
F-bx
2,400
1,450
1.15 1 1 1 0.9802 1 1     2,705
--
tension stress parallel to grain Ft 875 1.15 1 1             --
shear stress parallel to grain Fvx 210 1.15 1 1             242
compression stress perpendicular to grain Fc⊥x 740
650
  1 1           1 --
compression stress parallel to grain Fc 1,000 1.15 1 1         1   --
radial tension Frt -- 1.15 1 1             --
modulus of elasticity (or MOE) Ex 1,700,000   1 1             1,700,000
modulus of elasticity for stability calculations Ex min 880,000   1 1             --
modulus of elasticity (or MOE) Ey 1,300,000   1 1             1,300,000
modulus of elasticity for stability calculations Ey min 670,000   1 1             --

 

 

Actual Stresses (ASD) and Deflection

 

vertical ground reactions:  wL/2
Vmax = +/- wL/2
Mmax = wL2/8

fb = M/S = (500 lb/ft x 1 ft/12 in.)(20 ft x 12 in/ft)2 / (8)(308.8) = 971.5 psi

fv = 1.5V/A = 1.5 (500 lb/ft x 1 ft/12 in.)(20 ft x 12 in/ft) / 2(96.25 in.2) = 77.92 psi

ΔS = 5(300 lb/ft x 1 ft/12 in.)(20 ft x 12 in./ft)4 / 384(1,700,000 psi)(2972 in.4) = 0.2138 in.

ΔD+S = 5((200/2+300 lb/ft) x 1 ft/12 in.)(20 ft x 12 in./ft)4 / 384(1,700,000 psi)(2972 in.4) = 0.2850 in.  (dead load may be reduce by 1/2 if MC < 16% when installed and kept dry, IBC table 1604.3 footnote d)

 

Is Member Adequate? (ASD)

 

bending:  2,705 psi > 971.5 psi  ✓

shear:  242 psi > 77.92 psi  ✓

deflectionS:  0.6667 in. > 0.2138 in.  ✓

deflectionD+S:  1.000 in. > 0.2850 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)