1. Using Fy=50 ksi select the lightest W24 available for the service loads Pp = 300 k and Pr = 400 k. Lc= 8 ft. Use LRFD method only. 2. Select a suitable beam section for a simply supported beam by LRFD method. The span of the beam 20 ft. It carries a service dead load of 2 k/ft and service concentrated live load of 30 k at the center of the beam. Assume full lateral support. Use Fy = 50 ksi.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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1. Using Fy-50 ksi select the lightest W24 available for the service loads Pp = 300 k and Pr= 400 k
Lc= 8 ft. Use LRFD method only.
%3D
2. Select a suitable beam section for a simply supported beam by LRFD method. The span of the beam
20 ft. It carries a service dead load of 2 k/ft and service concentrated live load of 30 k at the center of
the beam. Assume full lateral support. Use Fy = 50 ksi.
3. Check the adequacy of the beam section of Q 2 for shear.
4. Check the adequacy of the beam of Q.2 from deflection criteria. The maximum permissible central
deflection is 1/360 of the span. Use the following formula (from AISC manual given) for deflection
calculation: I value is supplied in the last table.
ML?
A =
CI,
w (klft)
C, = 201
(b)
C, = 161
(a)
Transcribed Image Text:1. Using Fy-50 ksi select the lightest W24 available for the service loads Pp = 300 k and Pr= 400 k Lc= 8 ft. Use LRFD method only. %3D 2. Select a suitable beam section for a simply supported beam by LRFD method. The span of the beam 20 ft. It carries a service dead load of 2 k/ft and service concentrated live load of 30 k at the center of the beam. Assume full lateral support. Use Fy = 50 ksi. 3. Check the adequacy of the beam section of Q 2 for shear. 4. Check the adequacy of the beam of Q.2 from deflection criteria. The maximum permissible central deflection is 1/360 of the span. Use the following formula (from AISC manual given) for deflection calculation: I value is supplied in the last table. ML? A = CI, w (klft) C, = 201 (b) C, = 161 (a)
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