TMHL TMHL (Del I, teori; 1 p.) SOLUTION I. II.. III. Fig. 1.1

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TMHL6 204-08-30 (Del I, teori; p.). Fig.. shows three cases of sharp cracks in a sheet of metal. In all three cases, the sheet is assumed to be very large in comparison with the crack. Note the different crack lengths and temperatures! The of the material is temperature-dependent according to Fig..2. Order the three cases after risk of crack growth. Short mathematical explanations (but not detailed computations!) are required. I. II.. III. Fig.. Nr.. Highest risk of growth III I 3. Lowest risk of growth II Fig..2 Therefore: Compare = for the three cases; the higher, the more dangerous. Fall I: Fall II: Fall III:

TMHL6 204-08-30 (Del I, teori; p.) 2. By the Irwin theory, a plastic zone length can be computed. In plane stress, for instance, Cf Fig. 2, which is taken from the textbook. Describe how this theory can be used to set up a corrected crack length for computation in high-load cases (where the plastic zone size can be assumed to be non-negligible). Fig. 2 Use a corrected crack length received by adding the stretch, which is the stretch that the stress curve has moved horizontally in Fig. 2 in order to keep stress equilibrium. TMHL6 204-08-30 (Del I, teori; p.) 3. The load history in the figure is to be used for a fatigue analysis by a linear Palmgren-Miner theory. Using rainflow count, three load cycles can be identified. Which? (Explain by marking the drop paths in the figure!)

2 3 4 6 5 Drop table P i i b D op No. Range (MPa) -20 80 2 60 40 3 40 60 4 80-20 5 0 40 6 40 0 Cycle No. Drops Nos. Cycle (MPa) + 4-20 80-20 2 2 + 3 60 40 60 3 5 + 6 0 40 0 ANSWER TMHL6 204-08-30 (Del I, teori; p.) 4. Data for Paris w are frequently given in non-si units. is, for instance, often given so that inserted in will give in m. Show how such a value of is converted into an SI value such that i s d i P is w i will give in m.

With and according to the first case of the text: With and : TMHL6 204-08-30 (Del II, problem; 3 p.) 5. A beam containing a long, central crack ( and ) is loaded by an evenly distributed load. See figure! The material is linearly elastic with elastic modulus. Compute the energy release rate! Note! It is (as always) a good idea to start by identifying and using all symmetries. The solution can be done with an energy method or by the mathematical definition of the J integral. Both are shown here; it can be seen that the energy method gives the shortest way.. Alt. I Energy method: Horizontal and vertical symmetries the whole structure can be considered to consist of 4 equal beams. Study, for instance, the upper left beam: The end section force and moment at the right-hand end of it will be { Free-body diagram + moment equilibrium

We can now compute the strain energy of this beam: [ ] and the energy release rate becomes [ ] Alt II J integral method We therefore have

[ ] 2 We thus need. We can use linear beam deflection theory: We therefore have and 3 4 5 Conclusion

TMHL6 204-08-30 (Del II, problem; 3 p.) 6. A thin-walled cylindrical pressure vessel is loaded by a time-varying inner pressure { in the pressure vessel wall. The pressure vessel is to be designed against HCF, using the Sines hypothesis. Material parameters in the Sines hypothesis : and. Compute the necessary wall thickness, provided that is known. [ ] [ ] ( )

TMHL6 204-08-30 (Del II, problem; 3 p.) 7. A component, which for simplicity can be treated as a flat specimen with a rectangular cross section, has an area reduction with notch geometry; see Fig.. The load is a time-varying longitudinal force, which in the thinner right-hand half gives an evenly distributed stress. The component is to be designed for an LCF life of 5000 cycles. Compute the stress amplitude that can be allowed! Fig. Data: P Yta polerad/ surface polished MPa Ramberg-Osgood curve, see Fig. 2 Fig. 2

Morrow s equation Ramberg-Osgood Reading in the diagram of Fig. 2 gives. (An extremely careful plotting could possibly have given instead, but the relevance of such extreme detail is questionable; let us keep with 450 MPa.) Neuber We need Accurate reading in the table on top left of p. 64 gives { The Neuber equation then gives PS. Note that with arrived at the same answer! and, again, computing to 3 significant digits, we woluld have

TMHL6 204-08-30 (Del II, problem; 3 p.) 8. A long, thin-walled pressure vessel is loaded by a time-varying pressure that causes the tangential stress to vary as (everything in SI units). The pressure vessel will be inspected for cracks in -year intervals, i.e. every 8760 cycles. For safety, it has therefore been decided that it must be designed to stand 20 000 cycles between inspections. D si, T m i o ows P is w and since temperatures may be very low in parts of the vessel, a low design fracture toughness is also assumed, namely Compute the largest longitudinal crack that can then be allowed in the pressure vessel wall at the beginning of such a 20 000 cycle interval. Paris law ( ) [ ] where is the initial crack length, is the critical crack length and d si is the design life (between inspections). Solving for from Eq. () above gives [ ( ) ] To evaluate the expression in the right-hand member of Eq. (2), we need. is the crack length which gives static fracture at the maximum stress 60 MPa of a cycle:

Eq. (2) can now be evaluated with all data known: