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
NDS Supplement section 3.3CD = 0.9 (D), 1.25 (Lr)
Ct = 1 (given)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
NDS Supplement section 4.3Cb = 1 (not enough info given)
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
1.251 1 1 1 1 1 1 --
--
--tension stress parallel to grain Ft 600 0.9
1.251 1 1 1 --
--
--shear stress parallel to grain Fv 150 0.9
1.251 1 1 --
--
--compression stress perpendicular to grain Fc⊥ 405 1 1 1 1 -- compression stress parallel to grain Fc 1,550 0.9
1.251 1 1 0.3345, 0.2707, 0.2228
0.2485, 0.1994, 0.16321 466.6, 377.6, 310.8
481.5, 386.3, 316.2modulus 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