Question

structures homework, please answer all parts of the question thank you

2. Wood Beam Design Design a 2x dimensioned lumber floor joist to carry the given dead + live floor load (neglect joist selfweight). Assume the floor meets conditions of 4.4.1 so CL=1.0. Also Ct, Cfu, and Ci = 1.0. Find the short term deflection of your chosen beam under live load only (100% LL is short term). Compare your LL deflection with the code limit of L/360. SPAN DATASET: 1 HEM- Wood Species FIR Wood Grade No.1 Span 22 FT JOISTS Joist Spacing, o.c. Moisture Content, m.c. 16 IN 20 % 5 PSF Floor DL Floor LL 30 PSF Correct Answer Your Response 23 Question Score PSI SUBMIT Tabulated Allow. Bending Stress, Fb SUBMIT 2 Tabulated Allow. Shear Stress, Fv PSI Tabulated Modulus of Elasticity, E SUBMIT PSI SUBMIT Total Applied Floor Load, (DL+LL) PSF SUBMIT Load on Joist, w PLF Actual Beam Bending Moment, M FT-LB SUBMIT Actual Maximum Shear Force (at reaction), V SUBMIT LBS SUBMIT Nominal Depth of the Final Joist Used IN SUBMIT Size Factor, CF 10 Repetetive Member Factor, Cr SUBMIT SUBMIT 11 Wet Service Factor for Fb, CM_b Wet Service Factor for Fv. CM_v SUMT 12 13 Factored Allow. Bending Stress, F'b SUBMIT PSI 14 Factored Allow. Shear Stress, F'v PSI SUBMIT Actual Bending Stress, fb actual SUBMIT 15 PSI 16 Actual Shear Stress, fv_actual PSI SUBMIT SUBMIT 17 Factored Allow. Modulus of Elasticity, E PSI SUBMIT 18 Short Term Deflection for 100% LL IN SUBMIT 19 Short Term Deflection Limit for L/360 IN Deflection Passing: enter "1" for pass or "0" for 20 (1 or 0) SUBMIT fail 0/ 100 Current Score:

Answer 1

aood Hem-fir NO 1 GYA de Span of joist = 22 H=L Joist spacing Moisture content 20.(M.e) 16 IN (O.C) Y. (M.C) Ploor DL =5 PSF Floor LL = 30 PSF #em - fir No grade, from NDS Table 4A For O Tabulated Alora Bending stre 975 PSI %3D ) Tabulaled Allow shear <treys 75 PSI Modulus of EINtici ty Gabulated 1500000 PSI

Total Applied Floor LOnd (OL+L) (D2+LL)= 5+30 = 35 PSF O Lond on Joist (w) 3(3otst spacing) x (DL+LLI 16 X 35 46.7 PLF 12 46.7 PLF 4-ctual beam bending manent (M). 46.7x22 2825.35 FT-Lb 2825.35 FT-LB Shear oforee (at seaction) @. Actual Maxiimum 46.7x22 Vmax = 513.7 LBS 513.7 LBS Nominal Depth of the f inal Joist used . 6. (2)

nd For breadtr =2 IN dep thr .d= = 3 IN. 3 IN size Facton (C) From Table rf NDS

Repetative Member factor (Co) = 1-15 Service factor fr ,(CM-b) () wet (F) (C) = 975x 1S= 1462.5 > 115O PST = 0.85 wet service faetor fer f leM-v) = 0.97 Faetred Allow Bending 8tress (F). 13 6 = 1 (For Normal load diration) = LOad deuration faster G= ! (Fr continuery lateral support Beam stabiliky faeter 975 x 1.15X 1-51x 1 = 1681. 88 PST = 168188 PSI Strey (F) 9 Factored Allow Shear 4 = 2 for horiztal Sheas factor V. F= 150 PSI - 150 PSI

Aetual bending 8trey fo actuel 15 Bendling moment S = seetion moduly table 1B, for 2"x 3" fo For NDS S= S = 1.56 in" メ丈 2825. 35 FT-LB 1.56 IN 2825.35x 12 - 21738 5 PSI to-actual 16 Actual sheerr 8tress: Fr -actral 2(br4) From NDS able 1B, D = 2 5 IN b = 1:5 IN 2x 513-7 2 (1.5x 2.5) = 205.5 PSI 131 = 205.5 PSI aetual fretored Alloew Moduluy of Elasticity (E 1500000 PST

short term deflection ter 100y LL 18 A mar 384 EI I3 I=1:953 IN (Foom NDS table IB).

16 40 PLF 12 5x40 x 22 x ( 1728 in/Fr3) 384 x 1.SX10° x1953 IN. 72 term DeflectHiom 72 IN. Short (1) Short term deftectio liamit L/360 22X12 Alim = = 0.73 IN 3 GO 360 0.73 IN Alim Passing Deflection 20 Amax > Alim Fail, L0 o.