The steel bolt of cross-sectional area A0 is placed inside the aluminum tube, also of cross-sectional area A0. The assembly is completed by making the nut ‘‘fingertight.’’ The dimensions of the reduced segment of the bolt (length b and cross-sectional area A) are designed so that the segment will yield when the temperature of the assembly is increased by 200°F. Write an algorithm that determines the relationship between A/A0 and b/L that satisfies this design requirement. Plot A=A0 against b/L from b/L = 0 to 1.0. Use the properties of steel and aluminum shown in the figure.
The steel bolt of cross-sectional area A0 is placed inside the aluminum tube, also of cross-sectional area A0. The assembly is completed by making the nut ‘‘fingertight.’’ The dimensions of the reduced segment of the bolt (length b and cross-sectional area A) are designed so that the segment will yield when the temperature of the assembly is increased by 200°F. Write an algorithm that determines the relationship between A/A0 and b/L that satisfies this design requirement. Plot A=A0 against b/L from b/L = 0 to 1.0. Use the properties of steel and aluminum shown in the figure.
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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The steel bolt of cross-sectional area A0 is placed inside the aluminum tube, also of cross-sectional area A0. The assembly is completed by making the nut ‘‘fingertight.’’ The dimensions of the reduced segment of the bolt (length b and cross-sectional area A) are designed so that the segment will yield when the temperature of the assembly is increased by 200°F. Write an algorithm that determines the relationship between A/A0 and b/L that satisfies this design requirement. Plot A=A0 against b/L from b/L = 0 to 1.0. Use the properties of steel and aluminum shown in the figure.
Transcribed Image Text:Area A
%3D
Aluminum
Steel
Area A
E = 10 x 10 psi
a = 12.8 x 10-6/°F
Area = A,
E = 29 x 10* psi
* = 6.5 x 10-6F
p = 35 x 10' psi
%3D
%3D
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