The composite bar, firmly attached to unyielding supports, is initially stressfree. What maximum axial load P can be applied if the allowable stresses are 10 ksi for aluminum and 18 ksi for steel? Aluminum A = 1.25 in.² E = 10 x 106 psi -15 in.- Steel A = 2.0 in.2 E = 29 x 106 psi * P 12 in.-
The composite bar, firmly attached to unyielding supports, is initially stressfree. What maximum axial load P can be applied if the allowable stresses are 10 ksi for aluminum and 18 ksi for steel? Aluminum A = 1.25 in.² E = 10 x 106 psi -15 in.- Steel A = 2.0 in.2 E = 29 x 106 psi * P 12 in.-
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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Question
a. what is P to satisfy the condition in aluminum? in pounds, 2 decimal places
b. what is P to satisfy the condition in steel?
![The composite bar, firmly attached to unyielding supports, is
initially stressfree. What maximum axial load P can be
applied if the allowable stresses are 10 ksi for aluminum
and 18 ksi for steel?
Aluminum
A = 1.25 in.²
E = 10 x 106 psi
15 in.-
Steel
A = 2.0 in.2
E = 29 × 106 psi
-12 in.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9c78883-4a72-4499-a442-31d204b40d50%2Fdf94a5a6-f543-419d-ad20-c49603359463%2Fxdkhi0l_processed.png&w=3840&q=75)
Transcribed Image Text:The composite bar, firmly attached to unyielding supports, is
initially stressfree. What maximum axial load P can be
applied if the allowable stresses are 10 ksi for aluminum
and 18 ksi for steel?
Aluminum
A = 1.25 in.²
E = 10 x 106 psi
15 in.-
Steel
A = 2.0 in.2
E = 29 × 106 psi
-12 in.
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