Problem 7.20
Given
Member size sawn lumber Load axial compression
D = 20 kip
S = 55 kipStress grade and species No. 1 DF-L Unbraced length lu = 8, 10, 14, 18, 22 ft
same for both axesAdjustment factors Ke = 1
CM = 1
Ct = 1
Ci = 1
Load Combinations
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 20 kip D 3 16-10 D + H + F + (Lr or S or R) D + S 75 kip S 4 16-11 D + H + F + 0.75(L + T) + 0.75(Lr or S or R) D + 0.75S 61.25 kip S 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 + 0.75S 61.25 kip S 7 16-14 0.6D + W + H -- -- 8 16-15 0.6D + 0.7E + H -- -- (D)max = 20 kip
(D+S)max = 75 kip
Trial Size Based on (D+S) Load Combination
fc = P/A ≤ F'c
Rearranging,
A ≥ F'cP = P / (0.9 or 1.15)CFCPFc
where CF and CP depend on A, so let's try
A ≈ P / (0.9 or 1.15)Fc = 75,000 / (0.9 or 1.15)(1,500) = 55.6 or 43.5 in.2
where Fc = 1,500 psi (NDS Supplement table 4A) is for Dimension Lumber.
However, an 8x8 (assume a square cross section; A8x8 = 56.25 in.2) or larger (NDS Supplement table 1B) is apparently needed, which would be a Timber instead of Dimensional Lumber.
Adjusting for Posts and Timbers (NDS Supplement table 4D),
A ≈ P / (0.9 or 1.15)Fc = 75,000 / (0.9 or 1.15)(1,000) = 83.3 or 65.2 in.2
From NDS Supplement table 4A,
A8x8 = 56.25 in.2, Sxx = Syy = 70.31 in.3, Ixx = Iyy = 263.7 in.4
A10x10 = 90.25 in.2, Sxx = Syy = 142.9 in.3, Ixx = Iyy = 678.8 in.4
A12x12 = 132.3 in.2, Sxx = Syy = 253.5 in.3, Ixx = Iyy = 1458 in.4
Adjusted Design Values
NDS Supplement table 4D Cr = 1 (column)
CM = 1 (given)
CF = 1 (does not exceed 12")
Cfu = 1 (not a beam)NDS Supplement section 2.3
NDS Supplement section 3.3CD = 0.9 (D), 1.15 (S)
Ct = 1 (given)CL = 1 (no flexure)
NDS Supplement section 3.7
Ke = 1 (given)
Let l = lu (full length not given)
le = Ke l
le1/d1 = le2/d2 = Ke lu / (7.5 or 9.5 or 11.5 in.) (NDS Supplement table 1B)
FcE = 0.822 E'min / (le/d)2 = 0.822 (580,000 psi) / [Ke lu / (7.5 or 9.5 or 11.5 in.)]2
F*c = F'c without CP = 900 (D) or 1,150 (S) psi
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
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 4D)Adjustment Factors (Table 4.3.1) Adjusted Design
Values (psi)CD CM Ct CL CF Cfu Cr CP Ci CT Cb bending stress Fb 1,200 0.9
1.151 1 1 1 1 1 1 --
--
--tension stress parallel to grain Ft 825 0.9
1.151 1 1 1 --
--
--shear stress parallel to grain Fv 170 0.9
1.151 1 1 --
--
--compression stress perpendicular to grain Fc⊥ 625 1 1 1 1 -- compression stress parallel to grain Fc 1,000 0.9
1.151 1 1 see below 1 see below modulus of elasticity (or MOE) E 1,600,000 1 1 1 -- modulus of elasticity for stability calculations Emin 580,000 1 1 1 1 580,000
F*c = 900 (D) or 1,150 (S) psi
CP = (1+FcE/F*c)/2c - sqrt{[(1+FcE/F*c) / 2c]2 - (FcE/F*c)/c}
= {1+0.822 x 580,000 / [lu/(7.5 or 9.5 or 11.5 in.)]2F*c}/(2x0.8) - sqrt{[(1+0.822 x 580,000 / [lu/(7.5 or 9.5 or 11.5 in.)]2F*c) / (2x0.8)]2 - (0.822 x 580,000 / [lu/(7.5 or 9.5 or 11.5 in.)]2F*c)/0.8}calculations.xlsx
CP values 8x8 10x10 12x12 F*c (psi) F*c (psi) F*c (psi) 900 1,150 900 1,150 900 1,150 lu (in) 96 0.9257 0.9005 0.9567 0.9429 0.9715 0.9628 120 0.8727 0.8272 0.9280 0.9036 0.9535 0.9386 168 0.7094 0.6218 0.8367 0.7785 0.8981 0.8623 216 0.5236 0.4335 0.6993 0.6105 0.8101 0.7437 264 0.3807 0.3073 0.5508 0.4590 0.6928 0.6032
Actual Stress (ASD)
fc = P/A ≤ F'c
Pmax = F'cA = CPF*cA
(D)max = 20 kip
(D+S)max = 75 kipcalculations.xlsx
Pmax values
(lb)8x8 10x10 12x12 Load Combination Load Combination Load Combination D D + S D D + S D D + S lu (in) 96 46,865 58,251 77,710 97,866 115,636 146,433 120 44,181 53,512 75,376 93,786 113,492 142,747 168 35,913 40,222 67,959 80,797 106,902 131,149 216 26,506 28,041 56,805 63,365 96,428 113,101 264 19,274 19,880 44,742 47,635 82,456 91,745 ∴ An 8x8 would not work for any of the unbraced lengths. A 10x10 would work for lu = 8, 10 and 14 ft. A 12x12 would work for lu = 18 and 22 ft.
Considering Square and Non-Square Cross Sections
This spreadsheet contains additional calculations for non-square cross sections. Note that the le/d ratios used above for CP remained the same, because the unbraced length was the same for both axes and d therefore is the smaller dimension. The Fc and F*c values had to be modified for the cross sections that are considered beams-and-stringers instead of posts-and-timbers. The areas in the maximum-load calculation needed to be changed from A = d2 to A = d1d2.
The minimum column sizes, i.e. the columns with the smallest cross-sectional areas, that satisfy both the D and D+S loads are given in the following table.
lu
(ft)Minimum
SizeArea
(in.2)8 8x12 86.25 10 8x12 86.25 14 10x10 90.25 18 10x12 109.25 22 10x16 147.25