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Unformatted text preview: 025 . 4 m A A π = = = The crosssectional area of the 10mmdiameter segment is ( ) 2 2 3 40000 010 . 4 m A = = (b) The deformation in segment (2) is thus ( )( ) ( )( ) 9 2 2 2 2 2 10 * 100 6400 / 2 . 1 39000 = = E A L F mm 9534 . 2 = The deformation in segment (1) can be computed as ( )( ) ( )( ) 9 1 1 1 1 1 10 * 100 6400 / 8 . 1 79000 π δ = = E A L F mm 8969 . 2 1 = The deformation in segment (3) can be computed as ( )( ) ( )( ) 9 3 3 3 3 3 10 * 100 40000 / 6 . 1 25000 = = E A L F mm 0930 . 5 3 = (c) Since segments (1) and (2) have the same crosssectional area, the maximum normal stress in these two segments occurs where the axial force is greater; that is, in segment (1): 6400 / 79000 1 1 1 σ = = A F MPa 94 . 160 1 = The normal stress in segment (3) is 40000 / 25000 3 3 3 = = A F MPa 31 . 318 3 = Therefore, the maximum normal stress in the axial member occurs in segment (3): MPa 31 . 318 3 max = =...
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This note was uploaded on 02/08/2011 for the course ME 650:291 taught by Professor Weng during the Fall '11 term at Rutgers.
 Fall '11
 Weng

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