Problem 7.24

 

Given

 

Member size 2x4 stud wall 16 in. OC
Load axial compression for D and Lr
Stress grade and species Construction grade Hem-Fir
Unbraced length lu = 8, 9 or 10 ft for strong axis
lu = 0 for weak axis (sheathing)
Adjustment factors CM = 1
Ct = 1
Ci = 1

 

 

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 D
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 + 0.75Lr 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 = 5.25 in.2
Sxx = 3.063 in.3
Syy = 1.313 in.3
Ixx = 5.359 in.4
Iyy = 0.984 in.4

 

 

Adjusted Design Values

 

NDS Supplement table 4A

Cr = 1 (column)
 CM = 1 (given)
CF =  1 for all properties (construction)
Cfu = 1 (not a beam)

NDS Supplement section 2.3

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

NDS Supplement section 3.3

CL = 1 (no flexure)

NDS Supplement section 3.7

Ke = 1  (assumed)
le = Ke l
le1/d1 =  Ke lu1 / 3.5 in. = (96 or 108 or 120 in.) / 3.5 in. = 27.43, 30.86, or 34.29 (strong) 
le2/d2 =  Ke lu2 / 1.5 in. = 0 / 7.5 in. = 0 (weak)
 le1/d1 >  le2/d2  (strong axis governs)
 FcE = 0.822 E'min / (le/d)2 = 513.5, 405.7, or 328.6 psi
F*c = F'c without CP = 1,395 (D) or 1,938 (Lr) psi
c = 0.8
CP = (1+FcE/F*c)/2c - sqrt{[(1+FcE/F*c) / 2c]2 - (FcE/F*c)/c}
      = 0.3345, 0.2707, or 0.2228 (D)
      = 0.2485, 0.1994, or 0.1632 (Lr)

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 975 0.9
1.25		 1 1 1 1 1 1   1     --
--
--
tension stress parallel to grain Ft 600 0.9
1.25		 1 1   1       1     --
--
--
shear stress parallel to grain Fv 150 0.9
1.25		 1 1           1     --
--
--
compression stress perpendicular to grain Fc⊥ 405   1 1           1   1 --
compression stress parallel to grain Fc 1,550 0.9
1.25		 1 1   1     0.3345, 0.2707, 0.2228
0.2485, 0.1994, 0.1632		 1     466.6, 377.6, 310.8
481.5, 386.3, 316.2
modulus of elasticity (or MOE) E 1,300,000   1 1           1     --
modulus of elasticity for stability calculations Emin 470,000   1 1           1 1   470,000

 

 

Actual Stress (ASD)

 

fc = P/A ≤ F'c

Pmax = F'cA  

L (ft) Pmax (lb)
D D+Lr
8 2,450 2,528
9 1,982 2,028
10 1,632 1,660

wmax = Pmax / (16 in. x 1 ft/12 in.)

L (ft) wmax (lb/ft)
D D + Lr
8 1,838 1,896
9 1,487 1,521
10 1,224 1,245