Question

Select the lightest adequate W-section with a nominal depth of 14 in or less, for a 34ft. beam with a total uniformly distributed load of 1000 lb/ft. Select the beam based only on bending stress then check to see if deflection is O.K. LL Allowable deflection = L/360 TL Allowable deflection = L/240 Fb = 2/3 FY Fy = 50,000 lb/in2 LL is 70% of total load

M max = Select one: a. 4.25 k-ft b. 144.5 k-ft c. 101.1 k-ft d. 24.5 k-ft

Sr = Select one: a. 25 in³ b. 42.25 in³ c. 45 in³ d. 52 in³

Selection = Select one: a. W14x48 b. W18x35 c. W12x50 d. W21x44

Δ _{LL allow} = Select one: a. 2.14 in. b. 1.5 in. c. 1.13 in. d. 1.7 in

Δ _{LL max} = Select one: a. 2.14 in. b. 1.5 in. c. 1.13 in. d. 1.7 in.

Deflection O.K.?

Select one:

a. Yes

b. No

Answer 1

Given Datar nomind depth Less Length of beam (2) 34 ft. Total Uniform distributed Load 1000 Lolft DLL ) allow= 3uX12 - 1.13 in 360 360 D22) T may 34x12 느 240 1.7in 240 1000 lbft A tu 34' ☺) WL² 1000 + 34 11 ع امه 8. lu4.5x1000 lb-ft "max 144.5 Koft 2: M luy.5 X 12 Sr - 1 Fb 2 / 3 x 50 Sr= 52.02 in (iv) W12x50 for Sx= Select G4.2 in3 So an W 1250 Wlu x48 for Sx= in 3 To.2

5 wil 38G ET W 12 no 54. 391 inh F100 lb lift 3 Out of Total Load 0-7 kap lift 54 0-4x (34x1229 ܬܐ Ta 3818 29000 x 391 DLL 1.856 in >1-4 in (Not adequate ) SO go for w lux 48 Section W 14x18 =) 2184 iny ५ Dal 0.7 12 * (34412) 1.5 in 384 x 29000 x 484 <).fin. (ok). . "max luu.5 koft [" option ." ] Sr 52 in² [option "d" [ option "] section Wi4 X 48 ܐܐ0 allow = 1.13 in [option "c'] [option "d" DLL max lit in :. Deflection "OK"