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

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22.314/1.56/2.084/13.14 Fall 2006 Problem Set II Solution 1. Stress intensity = max {| σ r σ θ | , | σ θ σ z | , σ z σ r |} For thin wall approximation: σ r = P i + P o (1) 2 P i R 2 P o ( R + t ) 2 σ z = ( R + t ) 2 R 2 (2) σ θ = P i P o ( R + t ) (3) t 2 Therefore: S thin = σ θ σ r = P i P o ( R + t ) + P i + P o (4) t 2 2 Thick wall solution: Equilibrium in radial direction gives: r + σ r σ θ = 0 (5) dr r Hook’s law: 1 r = ( σ r νσ θ νσ z ) (6) E 1 θ = ( σ θ νσ r νσ z ) (7) E 1 z = ( σ z νσ r νσ θ ) (8) E du Since θ = u/r , r = dr , we get: d� θ 1 = ( r θ ) (9) dr r For this close end cylinder far from the end, plane stress condition is assumed, i.e., σ z is const. Plug Eq 6 and Eq 5into Eq 9, we get d ( σ θ + σ r ) = 0 (10) dr Plug Eq 10 into Eq 5, we get d 1 d ( r 2 σ r ) = 0 (11) dr r dr 1
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With B.C. σ r ( r = R ) = P i and σ r ( r = R + t ) = P o , we get: σ r = P i ( R r ) 2 + (1 ( R r ) 2 ) P o ( R + t ) 2 + P i R 2 ( R + t ) 2 R 2 σ θ = P i ( R r ) 2 + (1 + ( R r ) 2 ) P o ( R + t ) 2 + P i R 2 ( R + t ) 2 R 2 σ θ σ r = 2( R r ) 2
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