ing is trying to decide if they would like to use a composite or noncomposite de
ID: 1714701 • Letter: I
Question
ing is trying to decide if they would like to use a composite or noncomposite deck is 5" thick, wc-150 lb/ft, rc = 4 ksi, CMCE 2315 Elements of Structural Design-Steel Section 6453 4 (6.0 points) The owner of a building if they wou floor system for the beams in the floor system below. The concrete Fy - So ksi and the live load is 100 required to support the loads (flexural check only), assuming psf. Using LRFD, calculate the most economical W section (plain steel) full lateral bracing of the top compression flange ed. Using the same W-section, what is the design strength assuming a fully distance from the beam center line to the edge of slab 4 ft. Ignore self-weight and the beams are simply support composite beam? Assume the of beam Ginder Beams 24 ft 5. (4.0 points) Using LRFD, design the maximum-size SMAW fillet welds for the plates shown, if Pd - 50 k, Pl-75 k, and E70 electrodes are used. Hint: parameters for weld size are provided in Chapter J 2. P × 12 L=?Explanation / Answer
4) Tributary width of beam=7.5 ft
dead load=(5/12)*150=62.5 psf
live load = 100 psf
dead load on beam = 62.5*7.5=468.75 plf
live load on beam=100*7.5=750 plf
moment in beam due to dead load = 0.468*242/8=33.7 kip-ft
moment in beam due to live load = 0.75*242/8=54 kip-ft
factored moment in beam = (1.2*33.7)+(1.6*54)=126.84 kip-ft
plain steel wide flanged section sufficient from flexural capacity = W14x26 which has a flexural capapcity of 151 kip-ft
determination of composite capapcity:
effective flange width of beam = min(L/4,c/c spacing of beams) = min(6,7.5)=6ft=72"
Area of cross section of steel beam = 7.69in2
Total tension in beam=0.9*50*7.69=346.05 ikips
depth of compression in concrete =a=346.05/(0.85*72*4)=1.41"
depth of steel beam=13.91"
lever arm of beam = (13.91/2)+5-(1.41/2)= 11.25 "=0.9375 ft
moment capacity of full composite beam=346.05*0.9375=324.4 kip-ft
5)factored axial load = (1.2*50)+(1.6*75)=180 kips
given plate is 1/2" thick
maximum fillet weld size that can be used = (1/2)-(1/16)=7/16
let required length of weld be L
180=1.392*7*2L
L=9.23"
Provide 7/16 fillet weid of L=10"
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