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A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if D, = 2D,. 5. Solution: s, =65ksi N. = 8 F =1500 kips 3 7D _N¸F _ (8)(1500) A 65 D, = 8.85 in say 8-in D. =2D, = 2

A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if D, = 2D,. 5. Solution: s, =65ksi N. = 8 F =1500 kips 3 7D _N¸F _ (8)(1500) A 65 D, = 8.85 in say 8-in D. =2D, = 2

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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A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if D, = 2D,. 5. Solution: s, =65ksi N = 8 F =1500 kips 3D _N,F A= (8)(1500) 65 D, = 8.85 in say 8in D. =2D, = 2 8- |=172 in
Transcribed Image Text:A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if D, = 2D,. 5. Solution: s, =65ksi N = 8 F =1500 kips 3D _N,F A= (8)(1500) 65 D, = 8.85 in say 8in D. =2D, = 2 8- |=172 in
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