Question

this problem uses 2015 NDS supplement

PROBLEM #1 (CET 4450 and EVE 4450) A4 x 6 carries an axial compressive force caused by dead, live, and roof live loads. Lumber is No. 1 DF-L. CM = 1.0, C = 1.0, and C = 1.0. Find: The maximum column load for each LRFD load combination if the unbraced length of the member is a.) 3 ft b.) 6 ftc.) 9 ft d.)12 ft For a, b, c, and d please find le=Kelu: leldy, assuming Ke Fon* (ksi) Fc (ksi) Cp Fen (ksi) Pn' - Fen'

Answer 1

Column height t1,- 30t = 37122364 Le=kel - kel = 1xH, = 1x 36 = 364 Assume both ends are pinned le366 - 10.285 [needs to be less than 50 '. [ OS 3.7.4.3] Column stability factor 0.822x e'x mon > Fees (Les Let z 0.822 x 6.2% LOS (10.285) 4817019 pse FCE => Fot FC XCD XCMX Cg X Cg xci 1350 x1x1x1x1 To - 1350 PS! Chwan 2018 FCE [1+ CPE It fee Y Ft fee ( ree 2 Cswan - 2 Cswan CSwan CS Scanned with CamScanner

451209 و135 63,11 48۱۶۰۱۹ و5) و 135 26.8 26.8 0.337 Axial capacity Feta FoxCpx Cmin & CAT CGx Cix cp 125000953 جم - (26062 PSP Pi=fcxA - (26 00611225 p = لمو 15442065 (b) H=64 Ce = 1x 6x 1 = 1 le به 42 305 هوا هما ده مئ 2005 Column Stability factor fce 0.822 xewin کہ 6:27 » 2 008 از 0.5) rice Pst 7 مه 196

FLE 12040_0.892 1350 From '1t0892 1+0.892 (140 89 29 0.892 0.6 Cp 2x0.8 Cp- 0.65 Axial capacity fo =1350x0.65 877.67 pse P23 h x Areen =877.67 x 12.25 P2210751.556 C) H329fH cexlx 9x12 & 108 ce 108 3.5 -30.857 Needs to less than 50 Colemn stability facth FCE" 0.82% 6.2X10S (30.8592 535.496 (Ice

from o 535:49 26.33 64 - fCE Eca 350 Cp و0.36 +1 لا \t 009 9800 0:3144 0.8 9_x 0.8 023568 م : دور از وطنهمه عوام ۲۱ : (750 003568 PS1 108 هه Pa = fcx Aven کد:12 808ه ح ) وہ۹ dj H12 12 off و4) = 12 28) x\ = م ce ۹( (42 دان 144 05 د Need to be less than 50 Columa stability facta FEE کہ)6:28 0 : 821x (40 (4) FLE - stم 201068 CS Scanned with CamScanner

FCE 301.08 20:2230 1350 from I to.2230 0,2230 Cp2 exorfo] (40.9729? 2x0.8 (p = 0.2172 Axial capacity fd = 135080.2172 - 293-22 psi Porfo fo х Area 293.22% 12:25 Po z3591.99 lb CS Scanned with CamScanner