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3. The aluminum block has the rectangular cross-section and is subjected to an axial compressive force of 8 kips. The 1.5-in side changed its length to 1.50014 inches. Use E = 10 x 10° ksi. Calculate the new length of the 2-inch side. Answer: Lew = 2.00019 in 1.5 in 2 in. 8 kip- 8 kip 3 in.

3. The aluminum block has the rectangular cross-section and is subjected to an axial compressive force of 8 kips. The 1.5-in side changed its length to 1.50014 inches. Use E = 10 x 10° ksi. Calculate the new length of the 2-inch side. Answer: Lew = 2.00019 in 1.5 in 2 in. 8 kip- 8 kip 3 in.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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STRENGTH OF MATERIALS (SIMPLE STRAIN:Poisson’s Ratio: Biaxial and Triaxial Deformations)100% UPVOTE! PLEASE INCLUDE FBD IF NECESSARY. THANK YOU! GOD BLESS! 

3. The aluminum block has the rectangular cross-section and is subjected to an axial compressive force of 8 kips. The 1.5-in side changed its length to 1.50014 inches. Use E = 10 x 10° ksi. Calculate the new Answer: Lnew = 2.00019 in length of the 2-inch side. 1.5 in 2 in. 8 kip- 8 kip 3 in.
Transcribed Image Text:3. The aluminum block has the rectangular cross-section and is subjected to an axial compressive force of 8 kips. The 1.5-in side changed its length to 1.50014 inches. Use E = 10 x 10° ksi. Calculate the new Answer: Lnew = 2.00019 in length of the 2-inch side. 1.5 in 2 in. 8 kip- 8 kip 3 in.
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