Homework SourseDetermine an approximate allowable centrally located through
Determine an approximate allowable centrally located through-thickness crack size, 2a, in a very wide, thick plate of 2024-T3 and 7075-T6 aluminum alloys subjected to a nominal stress of 270 MPa. Repeat he calculation for nominal stresses equal to Syl2. What is the significance of the above results?
Solution
1. in the above calculations we observe that the allowable crack size 2a gets reduced with increasing stress.
2. the inspection methods for the plate gets stringent with increasing stress as detactable crack length decreases. Hence the cost .
3. For the same KIc value, with increasing Sy, allows higher load capacity
4. Fracture stress is not dependent on the the yield strength, but on the critical crack size. thus cracks can cause fractrue at very low nominal stress where as classical methods may indicate safety of the part.
| KIc is the critical SIF at plane strain condition . The problem given is for thick plate i.e for plane strain conditions. Wide plate means the a/b i.e the ratio of crack size to plate width will be very small and we take it as inifinite plate. The fracture will occur when =KIc/a. from this equation we shall compute crack length 2a. | |||
| Aluminum alloy | |||
| 2024-T3 | 7075-T6 | ||
| KIc | 26 | 26 | MPam |
| 270 | 270 | MPa | |
| a | 0.00295168 | 0.00295168 | m |
| 2a | 5.903360303 | 5.903360303 | mm |
| = yield strength | 455 | 493 | MPa |
| a | 0.001039379 | 0.000885326 | m |
| 2a | 2.07875844 | 1.770651046 | mm |
| 1. in the above calculations we observe that the allowable crack size 2a gets reduced with increasing stress. 4. Fracture stress is not dependent on the the yield strength, but on the critical crack size. thus cracks can cause fractrue at very low nominal stress where as classical methods may indicate safety of the part. |
