(a) Explanation Given information: The diameter ( d A B ) of portion AB is 1.5 in . The diameter ( d B C ) of portion BC is 1 in . The diameter ( d C D ) of portion CD is 1.5 in . The length ( L A B ) of portion AB is 2 in . The length ( L B C ) of portion AB is 3 in . The length ( L C D ) of portion AB is 2 in . The total deformation ( δ ) is 0.002 in . Calculation: Show the expression of deformation to find the area as shown below: δ = ∑ P i L i A i E i = P E ∑ L i A i P = E δ ( ∑ L i A i ) − 1 A i = π 4 d i 2 Find the area ( A A B ) of portion AB using the relation as follows: ( A A B ) = π 4 ( d A B 2 ) (1) Substitute 1.5 in . for d A B in Equation (1). ( A A B ) = π 4 ( 1.5 2 ) = 1.7671 Find the area ( A B C ) of portion BC using the relation as follows: ( A B C ) = π 4 ( d B C 2 ) (2) Substitute 1.0 in . for d B C in Equation (2) (b) To determine Find the deformation of the central portion BC .

7th Edition

ISBN: 8220100257063

Author: BEER

Publisher: YUZU

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Chapter 2, Problem 128RP

(a)

To determine

Find the load P when total deformation is 0.002 in.

(b)

To determine

Find the deformation of the central portion BC.

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Chapter 2, Problem 127RP

Chapter 2, Problem 129RP

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Chapter 2 Solutions

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Find the load P when total deformation is 0.002 in .

Solution Summary: The author shows the expression of deformation to find the area as shown below.