Problem 7.11

 

Given

 

Member size 4x4
Load axial compression from D, L, Lr
Stress grade and species No. 1 DF-L
Adjustment factors CM = 1
Ct = 1
Ci = 1
Unbraced length lu = 3, 6, 9, 12 ft

 

 

Load Combinations

 

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

 

 

Size (NDS Supplement table 1B)

 

A = 12.25 in.2
Sxx = Syy = 7.146 in.3
Ixx = Iyy = 12.51 in.4

 

Adjusted Design Values

 

NDS Supplement table 4A

Cr = 1 (not enough info given)
CM = 1 (given)
CF = 1.5 for Fb, 1.5 for Ft, 1.15 for Fc
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  (assumed; end conditions not given)
Let l = lu  (full length not given)
le = Ke l
le1/d1 = le2/d2 = Ke l / 3.5 in.
FcE = 0.822 E'min / (le/d)2
F*c = F'c without CP
c = 0.8
CP = (1+FcE/F*c)/2c - sqrt{[(1+FcE/F*c) / 2c]2 - (FcE/F*c)/c}

NDS Supplement section 3.10

Cb = 1 (not enough info given)

NDS Supplement section 4.3

Ci = 1 (given)

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 0.9
1.0
1.25		 1 1 1 1.5 1 1   1     --
--
--
tension stress parallel to grain Ft 675 0.9
1.0
1.25		 1 1   1.5       1     --
--
--
shear stress parallel to grain Fv 180 0.9
1.0
1.25		 1 1           1     --
--
--
compression stress perpendicular to grain Fc⊥ 625   1 1           1   1 --
compression stress parallel to grain Fc 1,500 0.9
1.0
1.25		 1 1   1.15     see below 1     see below
modulus of elasticity (or MOE) E 1,700,000   1 1           1     --
modulus of elasticity for stability calculations Emin 620,000   1 1           1 1   620,000

 

F*c = 1,553 (D), 1,725 (L), 2,156 (Lr) psi

CP = (1+FcE/F*c)/2c - sqrt{[(1+FcE/F*c) / 2c]2 - (FcE/F*c)/c}
      = (1+0.822 x 620,000 / (lu/3.5)2F*c)/(2x0.8) - sqrt{[(1+0.822 x 620,000 / (lu/3.5)2F*c) / (2x0.8)]2 - (0.822 x 620,000 / (lu/3.5)2F*c)/0.8}

CP values F*c (psi)
1,553 1,725 2,156
lu (in) 36 0.9220 0.9116 0.8842
72 0.5978 0.5576 0.4734
108 0.3156 0.2872 0.2340
144 0.1854 0.1678 0.1354
calculations.xlsx

 

 

Actual Stress (ASD)

 

fc = P/A ≤ F'c

Pmax = F'cA = CPF*cA

Pmax values
(lb)		 Load Combination
D D + L D + Lr or 
D + L + Lr
lu (in) 36 17,540 19,260 23,350
72 11,370 11,780 12,500
108 6,000 6,070 6,180
144 3,530 3,550 3,580
calculations.xlsx

Note that Pmax decreases as the unbraced length increases, for all of the load combinations.