6. (20 pts) A W16 x 36 steel (AS72, Grade 50) wide-flange beam is to span 24 ft
ID: 1712063 • Letter: 6
Question
6. (20 pts) A W16 x 36 steel (AS72, Grade 50) wide-flange beam is to span 24 ft on simple supports. The uperim posed dead load is w = 1.25 kips/n for the full span, meaning that it does not include the weight of the beam, in addition to a point dead load of 10 kips at midspan. This means that all loads are dead loads. Assuming its bending strength is adequate, you are required to use detailed calculation to show if this beam has adequate design bending strength, using LRFD method, comparing with its factored shear by considering only load combination 1 and 2.Explanation / Answer
Given beam is W16x36 of span 24 ft
Moment at midspan due to superimposed dead load = 1.25*242/8=90 kip-ft
Moment at misdpan due to point dead load = 10*24/4=60 kip-ft
Moment at midspan due to self weight = 0.036*242/8=2.592 kip-ft
Total service moment at midspan = 90+60+2.592=152.592 kip-ft
Design moment per load combination 1 =1.4D=1.4*152.592=213.63 kip-ft
Design moment per load combination 2=1.2D+1.6L = 1.2*152.592=183.11 kip-ft
Therefore, design moment = 213.63 kip-ft
Per Table 3-2 of AISC ,the moment capapcity of beam is 240 kip-ft (therefore bending capapcity is adequate)
Shear force at support due to superimposed dead load = 1.25*24/2=15 kips
Shear force at support due to point dead load = 10/2=5 kips
Shear force at support due to self weight = 0.036*24/2=0.432kip
Total service shear force at support = 15+5+.432=20.432kips
Design shear force due to load combination 1=1.4D=1.4*20.432=28.6 kips
Design shear force due to load combination 2=1.2D+1.6L = 1.2*20.432=24.52 kips
Design shear force = 28.6 kips
Shear strength of beam = 140 kips
Therefore, beam is adequate in shear
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.