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A W21x93 wide flange structural steel member, shown in Fig. 1, is a part of a braced frame. The member is laterally braced (braced about weak axis bending) at the supports and at midpoint (B). Check whether a W21x93 section is adequate to support the loadings shown in Fig.1. The Load Combinations (L.C.s) are: L.C. = 1.2D + 1.0 E + 0.5L and/or L.C.2- 1.2D + 1.6L PD = 10k PL=6K WD = 2K/h WI 2Kt PD = 125k PL= 65K PD = 125K P=65K 5ft 5ft 5ft 5ft Mg= 145 Mg= 1855 PE = 115k PE =115K 5ft 5ft 5ft 5ft Fig. 1

A W21x93 wide flange structural steel member, shown in Fig. 1, is a part of a braced frame. The member is laterally braced (braced about weak axis bending) at the supports and at midpoint (B). Check whether a W21x93 section is adequate to support the loadings shown in Fig.1. The Load Combinations (L.C.s) are: L.C. = 1.2D + 1.0 E + 0.5L and/or L.C.2- 1.2D + 1.6L PD = 10k PL=6K WD = 2K/h WI 2Kt PD = 125k PL= 65K PD = 125K P=65K 5ft 5ft 5ft 5ft Mg= 145 Mg= 1855 PE = 115k PE =115K 5ft 5ft 5ft 5ft Fig. 1

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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A W21x93 wide flange structural steel member, shown in Fig. 1, is a part of a braced frame. The member is laterally braced (braced about weak axis bending) at the supports and at midpoint (B). Check whether a W21x93 section is adequate to support the loadings shown in Fig.1. The Load Combinations (L.C.s) are: L.C. = 1.2D + 1.0E + 0.5L and/or L.C.= 1.2D + 1.6L PD = 10K PL=6K WD = WI =2Kn PD = 125K - 65K %3D PD = 125K P 65K 5ft 5ft Sft 5ft Mg= 145k ME= 185K- PE = 115K B PE = 115K 5ft 5ft 5ft 5ft Fig. 1
Transcribed Image Text:A W21x93 wide flange structural steel member, shown in Fig. 1, is a part of a braced frame. The member is laterally braced (braced about weak axis bending) at the supports and at midpoint (B). Check whether a W21x93 section is adequate to support the loadings shown in Fig.1. The Load Combinations (L.C.s) are: L.C. = 1.2D + 1.0E + 0.5L and/or L.C.= 1.2D + 1.6L PD = 10K PL=6K WD = WI =2Kn PD = 125K - 65K %3D PD = 125K P 65K 5ft 5ft Sft 5ft Mg= 145k ME= 185K- PE = 115K B PE = 115K 5ft 5ft 5ft 5ft Fig. 1
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