problem2_sol - 22.314/1.56/2.084/13.14 Fall 2006 Problem...

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Unformatted text preview: 22.314/1.56/2.084/13.14 Fall 2006 Problem Set II Solution 1. Stress intensity = max {| r | , | z | , z r |} For thin wall approximation: r = P i + P o (1) 2 P i R 2 P o ( R + t ) 2 z = ( R + t ) 2 R 2 (2) = P i P o ( R + t ) (3) t 2 Therefore: S thin = r = P i P o ( R + t ) + P i + P o (4) t 2 2 Thick wall solution: Equilibrium in radial direction gives: d r + r = 0 (5) dr r Hooks law: 1 r = ( r z ) (6) E 1 = ( r z ) (7) E 1 z = ( z r ) (8) E du Since = u/r , r = dr , we get: d 1 = ( r ) (9) dr r For this close end cylinder far from the end, plane stress condition is assumed, i.e., z is const. Plug Eq 6 and Eq 5into Eq 9, we get d ( + r ) = 0 (10) dr Plug Eq 10 into Eq 5, we get d 1 d ( r 2 r ) = 0 (11) dr r dr 1 With B.C. r ( r = R ) = P i and r ( r = R + t ) = P o , we...
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problem2_sol - 22.314/1.56/2.084/13.14 Fall 2006 Problem...

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