Problem 7.17

 

Given

 

Member size 8-3/4 x 15
Axial loads D = 20 kip
L = 90 kip
Lr = 40 kip
Stress grade and species axial combination 2 DF glulam without special tension lams
Lengths L1 = 22 ft
L2 = 10 ft
L3 = 12 ft
Adjustment factors Ke = 1
CM = 1
Ct = 1
Unbraced length lu = L for strong axis
lu = L1 and L2 for weak axis

 

 

Load Combinations

 

Load Symbol Given Value
dead load D 20 kip
weight of ice Di 0
earthquake load E 0
load due to fluids F 0
flood loads Fa 0
load due to sub-surface pressure H 0
live load L 90 kip
roof live load Lr 40 kip
rain load R 0
snow load S 0
self-straining force T 0
wind load W 0
wind-on-ice Wi 0

ASCE 7 section 2.4

ASCE 7 IBC ASD Load Combination Summary Load Shortest-Duration Load
1 16-8 D + F D 20 kip D
2 16-9 D + H + F + L + T D + L 110 kip L
3 16-10 D + H + F + (Lr or S or R) D + Lr 60 kip Lr
4 16-11 D + H + F + 0.75(L + T) + 0.75(Lr or S or R) D + 0.75L + 0.75Lr 117.5 kip Lr
5 16-12 D + H + F + (W or 0.7E) D 20 kip D
6 16-13 D + H + F + 0.75(W or 0.7E) + 0.75L + 0.75(Lr or S or R) D + 0.75Lr 50 kip Lr
7 16-14 0.6D + W + H -- -- --
8 16-15 0.6D + 0.7E + H -- -- --

(D)max = 20 kip
(D+L)max = 110 kip
(D+Lr)max = 60 kip
(D+L+Lr)max = 117.5 kip

 

 

Size

 

A = 8-3/4 x 15 = 131.25 in.2

 

 

Adjusted Design Values

 

NDS Supplement table 5B

CV = (21/L)1/x (12/d)1/x (5.125/b)1/x 
      = (21/10?)1/10 (12/15)1/10 (5.125/8.75)1/10 = 0.??? ≤ 1.0
CM = 1 (given)
Cfu = 1 (not a beam)

NDS Supplement section 2.3

CD = 0.9 (D), 1.0 (L), 1.25 (Lr)
Ct = 1 (given)

NDS Supplement section 3.3

CL = 1 (no flexure)

NDS Supplement section 3.7

Ke = 1  (given)
le = Ke l
le1/d1 = Ke (L2 or L3) / 8.75 in. = 1 x 144 in. / 8.75 in. = 16.46  (L3 gives bigger number)
le2/d2 = Ke L1 / 15 in. = 1 x 264 in. / 15 in. = 17.60  (biggest)
le/d = 17.60
FcE = 0.822 E'min / (le/d)2 = 2,203 psi
F*c = F'c without CP = 1,755 (D) or 1,950 (L) or 2,438 (Lr) psi
c = 0.9
CP = (1+FcE/F*c)/2c - sqrt{[(1+FcE/F*c) / 2c]2 - (FcE/F*c)/c}
      = (1+2,203/F*c)/(2x0.9) - sqrt{[(1+2,203/F*c) / (2x0.9)]2 - (2,203/F*c)/0.9}
      = 0.8345 (D) or 0.8028 (L) or 0.7193 (Lr)

The solutions manual shows CP = 0.763 (D) or 0.731 (L) or 0.655 (Lr), but I can't figure out why.

NDS Supplement section 3.10

Cb = 1 (not enough info given)

 

Property Reference Design
Values (psi)
(Table 5B)		 Adjustment Factors (Table 4.3.1) Adjusted Design
Values (psi)
CD CM Ct CL CV Cfu Cc CP Cb
bending stress Fb 1,450 / 1,700 0.9
1.0
1.25		 1 1 1 ??? 1 1     --
--
--
tension stress parallel to grain Ft 1,250 0.9
1.0
1.25		 1 1             --
--
--
shear stress parallel to grain Fv 230 / 265 0.9
1.0
1.25		 1 1             --
--
--
compression stress perpendicular to grain Fc⊥ 560   1 1           1 --
compression stress parallel to grain Fc 1,950 0.9
1.0
1.25		 1 1         0.8345
0.8028
0.7193		   1,465
1,565
1,753
modulus of elasticity E 1,600,000   1 1             --
modulus of elasticity for stability calculations Emin 830,000   1 1             830,000

 

 

Actual Stress (ASD)

 

fc = P/A ≤ F'c

(D)max = 20 kip

fc = 20,000 lb / 131.25 in.2 = 152 psi ≤ 1,465 psi  ✓

(D+L)max = 110 kip

fc = 110,000 lb / 131.25 in.2 = 838 psi ≤ 1,565 psi  ✓

(D+Lr)max = 60 kip

fc = 60,000 lb / 131.25 in.2 = 457 psi ≤ 1,753 psi  ✓

(D+L+Lr)max = 117.5 kip

fc = 117,500 lb / 131.25 in.2 = 895 psi ≤ 1,753 psi  ✓