Assignment10-solution

Assignment10-solution - Problem 6.11. Compare the plastic...

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Problem 6.11. Compare the plastic zone sizes for plastic strain mode I fracture at failure in Al 2024-651 and Al 7075-651. Use loading where the stress intensity factor is equal to the fracture toughness (this is the maximum zone size). The plastic zone radius for plane strain mode 1 fracture is given by Eq. 6.50,    2 2 2 12 2 I p Y K r . Assume = 0.3 for both Al materials. For a given loading, the K I is fixed, so the radii will depend only on the yield stress. From Table 6.1, for 2024-651 the yield stress is 415MPa, and for 7075-651 it is 505MPa. This means that the plastic zone is smaller for 7075-651 by the ratio of (415/505) 2 =0.675. On the other hand, the fracture toughness of 2024-651 is 24MPa m, while for 7075-651 it is 29MPa m. At these values the radii are   2 2 2 5 2 2024 2 2 2 5 2 7075 24 1 2 0.3 8.52 10 0.0852 24 1 5 29 1 2 0.3 8.40 10 0.0840 25 0 5 p p rm m m m m Relative error is 1.4% Problem 6.13 : The split beam of Fig. 6.10 is subjected to a pair of cyclic opening forces P with N P 2000 max and 0 min P .
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Assignment10-solution - Problem 6.11. Compare the plastic...

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