{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment10-solution

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

This preview shows pages 1–2. Sign up to view the full content.

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 1 2 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 2 415 29 1 2 0.3 8.40 10 0.0840 2 505 p p r m mm r m mm Relative error is 1.4% Problem 6.13 : The split beam of Fig. 6.10 is subjected to a pair of cyclic opening forces

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online