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2.32 The rectangular block of material of length L and cross-sectional area A fits snugly between two rigid, lubricated walls. Derive the expression for the change in length of the block due to the axial load P. 0.75 in. 0.75 in.- Steel Plastic 10 in. P 4 in. Area = A Top view FIG. P2.32 FIG. P2.33

2.32 The rectangular block of material of length L and cross-sectional area A fits snugly between two rigid, lubricated walls. Derive the expression for the change in length of the block due to the axial load P. 0.75 in. 0.75 in.- Steel Plastic 10 in. P 4 in. Area = A Top view FIG. P2.32 FIG. P2.33

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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Using formula of General hooke's law

2.32 The rectangular block of material of length L and cross-sectional area A fits snugly between two rigid, lubricated walls. Derive the expression for the change in length of the block due to the axial load P. 0.75 in. - 0.75 in. Steel Plastic 10 in. P 14 4 in. Area = A- Top view FIG. P2.32 FIG. P2.33
Transcribed Image Text:2.32 The rectangular block of material of length L and cross-sectional area A fits snugly between two rigid, lubricated walls. Derive the expression for the change in length of the block due to the axial load P. 0.75 in. - 0.75 in. Steel Plastic 10 in. P 14 4 in. Area = A- Top view FIG. P2.32 FIG. P2.33
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