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Ierofanga [76]
3 years ago
15

A simply supported beam spans 25 ft and carries a uniformly distributed dead load of 0.6 kip/ft, including the beam self-weight,

and a live load of 2.1 kip/ft. Determine the minimum required plastic section modulus and select the lightest-weight W-shape to carry the moment. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD
Engineering
1 answer:
USPshnik [31]3 years ago
5 0

Answer:

Check the explanation

Explanation:

beam span = 25 ft

dead load = 0.6 kip/ft

live load = 2.1 kip/ft

factored load = 1.2*0.6 +1.6*2.1=4.08 kip/ft

moment in beam = 4.08*252/8=318.75 kip-ft = 3825 kip-in

design strength =0.9* 50 = 45 ksi

plastic section modulus required = 3825/45=85 in3

Moment in beam in ASD = (0.6+2.1)*252/8 = 210.9 kip-ft

lighest W section from LRFD = W21x44

lightest W section from ASD = W21x44

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