HW+9+solution (1)

# 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, ? ?Â¡ = ( âˆ† ) ? From the graph, picking points for ? ?Â¡ = 10 âˆ’ 4 and ? ?Â¡ = 10 âˆ’ 5 The corresponding ( Î” 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 Ïƒ o = P max /(2(b-a)t) where Ïƒ o = 1225 MPa From this, a = 29.84mm Since we do not know the crack length, assume F=1 K IC = FS max âˆšÏ€ 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) Î± F K IC (MPaâˆšm) 77.77 2.05 Î± 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 âˆ’ ? 2 âˆ’ ? 1 âˆ’ ? 2 âˆ†Â¡Â¢ Â£ ? Â¤ 1 âˆ’ ? 2 Â¥ C = 1.095 x 10-12 m/cycle (Mpa âˆš m)-m , m = 3.24 Here, Î” 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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