HW+9+solution (1) - MechEng 382 Winter 2010 Homework #9...

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Unformatted text preview: MechEng 382 Winter 2010 Homework #9 Solutions 1. Dowling 11.1 For the line is figure 11.3, ? ? = ( ) ? From the graph, picking points for ? ? = 10 4 and ? ? = 10 5 The corresponding ( K) points are 30 and 11. Solving the two equations for C and m, we get: C = 4 x 10-8 ; m = 2.3; MechEng 382 Winter 2010 Homework#9 Solutions 2 2. Dowling 11.37 (a) Crack length at failure For fully plastic yielding to occur o = P max /(2(b-a)t) where o = 1225 MPa From this, a = 29.84mm Since we do not know the crack length, assume F=1 K IC = FS max a where S max = P max /(2bt) = 263 MPa From this a = 77.77mm To refine a by trial and error, we calculate the corresponding F and K IC : a (mm) F K IC (MPam) 77.77 2.05 is too high! 30.00 0.79 1.76 142.10 29.00 0.76 1.65 130.98 From these iterations, it is clear that the plate will fail from fracture at a crack-length slightly <29mm. (b) Eqn 11.32: ? ?? = ? 1 ? 2 ? 1 ? 2 ? 1 ? 2 C = 1.095 x 10-12 m/cycle (Mpa m)-m , m = 3.24 Here, S = (P max- P min )/(2bt) = 175.44 MPa and all other values are given. Using F = 1.1 (for an F closer to F i than F f ), N if = 576,909 cycles This is not accurate because F is approximated to be a constant....
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HW+9+solution (1) - MechEng 382 Winter 2010 Homework #9...

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