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1. Calculate the weight (in pounds) of a 1-ft length of a steel built-up member having a cross section as shown. (unit weight of steel: 490 lb/ft³ ) (1ft =12 in. ) -R9 X T6 W16 X 67 L4 X 3 x L3 X 3 x }

1. Calculate the weight (in pounds) of a 1-ft length of a steel built-up member having a cross section as shown. (unit weight of steel: 490 lb/ft³ ) (1ft =12 in. ) -R9 X T6 W16 X 67 L4 X 3 x L3 X 3 x }

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Appendix Expressions for axially loaded compression member: KI For-< C. = 2T²E [1 – (Kl/r)²/(2C? )]Fy F, = 5 3(Kl/r) _ (Kl/r)³ 8Cc Fy |21²E 1212E KI For-> Cc Fa Fy 23(KI/r)2 Pa = FaA Expressions for beam: MR = F,S, in which S =- Expressions for tension member: Pt = A,F , in which F = 0.6Fy P; = A,Ft , in which F = 0.5Fu P = A.Ft , in which Ae = U An, F = 0.5Fu P = A,F, + A,F; , in which Fy = 0.30Fµ , F; = 0.50Fu P = nr,, in which r, = ApFv, ry = d t F, TABLE 2-1 Values for Reduction Coefficient, U U = 0.90 W, M, S shapes or their tees. Connection is to the flanges. Minimum of three bolts per line in the direction of Case I stress. (min.) (min.) U = 0,85 All shapes and built-up cross sections not meeting the requirements of case I. Minimum of three bolts per line in the direction of stress. Case II U = 0.75 All members whose connections have only two bolts per line in the direction of stress. Case III Units: ksi : kips per square inch kips: kilo pounds lb/ft: pounds per foot kips/ft: kilo pounds per foot
Transcribed Image Text:Appendix Expressions for axially loaded compression member: KI For-< C. = 2T²E [1 – (Kl/r)²/(2C? )]Fy F, = 5 3(Kl/r) _ (Kl/r)³ 8Cc Fy |21²E 1212E KI For-> Cc Fa Fy 23(KI/r)2 Pa = FaA Expressions for beam: MR = F,S, in which S =- Expressions for tension member: Pt = A,F , in which F = 0.6Fy P; = A,Ft , in which F = 0.5Fu P = A.Ft , in which Ae = U An, F = 0.5Fu P = A,F, + A,F; , in which Fy = 0.30Fµ , F; = 0.50Fu P = nr,, in which r, = ApFv, ry = d t F, TABLE 2-1 Values for Reduction Coefficient, U U = 0.90 W, M, S shapes or their tees. Connection is to the flanges. Minimum of three bolts per line in the direction of Case I stress. (min.) (min.) U = 0,85 All shapes and built-up cross sections not meeting the requirements of case I. Minimum of three bolts per line in the direction of stress. Case II U = 0.75 All members whose connections have only two bolts per line in the direction of stress. Case III Units: ksi : kips per square inch kips: kilo pounds lb/ft: pounds per foot kips/ft: kilo pounds per foot
1. Calculate the weight (in pounds) of a 1-ft length of a steel built-up member having a cross section as shown. (unit weight of steel: 490 lb/ft³ ) (1ft =12 in. ) 91 x 6a- W16 X 67 wL4X 3 x L3 X 3 x
Transcribed Image Text:1. Calculate the weight (in pounds) of a 1-ft length of a steel built-up member having a cross section as shown. (unit weight of steel: 490 lb/ft³ ) (1ft =12 in. ) 91 x 6a- W16 X 67 wL4X 3 x L3 X 3 x
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