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! Required information A 1-in.-square aluminum strut is maintained in the position shown by a pin support at A and by sets of rollers at B and C that prevent rotation of the strut in the plane of the figure. Given: LAB= 3.4 ft C B ab LCD T LBC T LAB Determine the magnitude of the allowable load. Consider only buckling in the plane of the figure and use E= 10.4 x 106 psi. Use a factor of safety of 3.2. (Round the final answer to two decimal places.)

! Required information A 1-in.-square aluminum strut is maintained in the position shown by a pin support at A and by sets of rollers at B and C that prevent rotation of the strut in the plane of the figure. Given: LAB= 3.4 ft C B ab LCD T LBC T LAB Determine the magnitude of the allowable load. Consider only buckling in the plane of the figure and use E= 10.4 x 106 psi. Use a factor of safety of 3.2. (Round the final answer to two decimal places.)

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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! Required information A 1-in.-square aluminum strut is maintained in the position shown by a pin support at A and by sets of rollers at B and C that prevent rotation of the strut in the plane of the figure. Given: LAB = 3.4 ft C B D P 000 LCD T LBC ↑ LAB Determine the magnitude of the allowable load. Consider only buckling in the plane of the figure and use E= 10.4 × 106 psi. Use a factor of safety of 3.2. (Round the final answer to two decimal places.)
Transcribed Image Text:! Required information A 1-in.-square aluminum strut is maintained in the position shown by a pin support at A and by sets of rollers at B and C that prevent rotation of the strut in the plane of the figure. Given: LAB = 3.4 ft C B D P 000 LCD T LBC ↑ LAB Determine the magnitude of the allowable load. Consider only buckling in the plane of the figure and use E= 10.4 × 106 psi. Use a factor of safety of 3.2. (Round the final answer to two decimal places.)
Explanation 12PcL² л²E d4 = - 2 12(13.0507x10³ N) ([A(1)] m)² π² (70×10⁹ Pa) d = [A(4)] × 10-³ m d= [A(4)] mm = [A(3)] × 10-⁹ m4
Transcribed Image Text:Explanation 12PcL² л²E d4 = - 2 12(13.0507x10³ N) ([A(1)] m)² π² (70×10⁹ Pa) d = [A(4)] × 10-³ m d= [A(4)] mm = [A(3)] × 10-⁹ m4
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