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1. You are to design an axially loaded member that is to carry a load of 5,000 lb. The member is to be made of ASTM A36 steel and have a total length of 24 in. Due to the design requirements the maximum allowable elongation of this member under the stated load is 0.025 in. Find the smallest allowable diameter of the member. a. Given: PL where ô is member elongation, P is axial load, L is initial member length, E is the material's Modulus of Elasticity, EA and A is the member's cross section area E136 = 29*10° psi (Beer, 746)

1. You are to design an axially loaded member that is to carry a load of 5,000 lb. The member is to be made of ASTM A36 steel and have a total length of 24 in. Due to the design requirements the maximum allowable elongation of this member under the stated load is 0.025 in. Find the smallest allowable diameter of the member. a. Given: PL where ô is member elongation, P is axial load, L is initial member length, E is the material's Modulus of Elasticity, EA and A is the member's cross section area E136 = 29*10° psi (Beer, 746)

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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1. You are to design an axially loaded member that is to carry a load of 5,000 lb. The member is to be made of ASTM A36 steel and have a total length of 24 in. Due to the design requirements the maximum allowable elongation of this member under the stated load is 0.025 in. Find the smallest allowable diameter of the member. a. Given: PL where ô is member elongation, P is axial load, L is initial member length, E is the material's Modulus of Elasticity, EA and A is the member's cross section area E136 = 29*10° psi (Beer, 746)
Transcribed Image Text:1. You are to design an axially loaded member that is to carry a load of 5,000 lb. The member is to be made of ASTM A36 steel and have a total length of 24 in. Due to the design requirements the maximum allowable elongation of this member under the stated load is 0.025 in. Find the smallest allowable diameter of the member. a. Given: PL where ô is member elongation, P is axial load, L is initial member length, E is the material's Modulus of Elasticity, EA and A is the member's cross section area E136 = 29*10° psi (Beer, 746)
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