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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

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.

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