Problem 02 Axial compression loads in the first interior column of the structure shown are: 95 kN (dead load), 50 kN (live load), and 75kN (wind Load). If the effective length factor k=0.65, determine the remaining allowance for moments in the column. Express in %. Use LRFD and A-50 steel (Fy=345 MPa, F₂ = 428 MPa). Radius of gyration is 82.6 mm and gross cross-sectional area is 36100 mm², fl-0.5. 7.5 m 7.8 m H 4.3 m 4.3 m

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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Problem 02
Axial compression loads in the first interior column of the structure shown are: 95 kN (dead load), 50 kN (live load), and 75kN (wind Load). If the
effective length factor k = 0.65, determine the remaining allowance for moments in the column. Express in %. Use LRFD and A-50 steel (Fy-345
MPa, F₁ = 428 MPa). Radius of gyration is 82.6 mm and gross cross-sectional area is 36100 mm², fl-0.5.
7.5 m
7.8 m
4.3 m
4.3 m
Transcribed Image Text:Problem 02 Axial compression loads in the first interior column of the structure shown are: 95 kN (dead load), 50 kN (live load), and 75kN (wind Load). If the effective length factor k = 0.65, determine the remaining allowance for moments in the column. Express in %. Use LRFD and A-50 steel (Fy-345 MPa, F₁ = 428 MPa). Radius of gyration is 82.6 mm and gross cross-sectional area is 36100 mm², fl-0.5. 7.5 m 7.8 m 4.3 m 4.3 m
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