Problem 7.20

 

Given

 

Member size	sawn lumber
Load	axial compression D = 20 kip S = 55 kip
Stress grade and species	No. 1 DF-L
Unbraced length	lu = 8, 10, 14, 18, 22 ft same for both axes
Adjustment 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

 

f_c = P/A ≤ F'_c

Rearranging,

A ≥ F'_cP = P / (0.9 or 1.15)C_FC_PF_c

where C_F and C_P depend on A, so let's try 

A ≈ P / (0.9 or 1.15)F_c = 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; A_8x8 = 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)F_c = 75,000 / (0.9 or 1.15)(1,000) = 83.3 or 65.2 in.²

From NDS Supplement table 4A,

A_8x8 = 56.25 in.², S_xx = S_yy = 70.31 in.³, I_xx = I_yy = 263.7 in.⁴
A_10x10 = 90.25 in.², S_xx = S_yy = 142.9 in.³, I_xx = I_yy = 678.8 in.⁴
A_12x12 = 132.3 in.², S_xx = S_yy = 253.5 in.³, I_xx = I_yy = 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 CD = 0.9 (D), 1.15 (S) Ct = 1 (given) NDS Supplement section 3.3 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 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 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 1.15	1	1	1	1	1	1		1			-- -- --
tension stress parallel to grain	Ft	825	0.9 1.15	1	1		1				1			-- -- --
shear stress parallel to grain	Fv	170	0.9 1.15	1	1						1			-- -- --
compression stress perpendicular to grain	Fc⊥	625		1	1						1		1	--
compression stress parallel to grain	Fc	1,000	0.9 1.15	1	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

C_P = (1+F_cE/F^*_c)/2c - sqrt{[(1+F_cE/F^*_c) / 2c]² - (F_cE/F^*_c)/c}
      = {1+0.822 x 580,000 / [l_u/(7.5 or 9.5 or 11.5 in.)]²F^*_c}/(2x0.8) - sqrt{[(1+0.822 x 580,000 / [l_u/(7.5 or 9.5 or 11.5 in.)]²F^*_c) / (2x0.8)]² - (0.822 x 580,000 / [l_u/(7.5 or 9.5 or 11.5 in.)]²F^*_c)/0.8}

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

calculations.xlsx

 

 

Actual Stress (ASD)

 

f_c = P/A ≤ F'_c

P_max = F'_cA = C_PF^*_cA

(D)_max = 20 kip
(D+S)_max = 75 kip

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

calculations.xlsx

∴ An 8x8 would not work for any of the unbraced lengths.  A 10x10 would work for l_u = 8, 10 and 14 ft.  A 12x12 would work for l_u = 18 and 22 ft.

 

 

Considering Square and Non-Square Cross Sections

 

This spreadsheet contains additional calculations for non-square cross sections.  Note that the l_e/d ratios used above for C_P remained the same, because the unbraced length was the same for both axes and d therefore is the smaller dimension.  The F_c 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 Size	Area (in.2)
8	8x12	86.25
10	8x12	86.25
14	10x10	90.25
18	10x12	109.25
22	10x16	147.25