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Unformatted text preview: AE 321 – Solution to Practice Problems Chapter 7: Failure and Fatigue 1a. We start with the following trial solution. [ ] ! ! ! " # $ $ $ % & ’ ’ = P P ( Note that the entries in the upper left of the stress tensor, i.e. 21 12 22 11 , , ! ! ! ! = , arise from the inplane hydrostatic case, and that all other entries are zero because the cylinder is tractionfree in the x 3direction. It can be easily shown that equilibrium is satisfied since P is constant. Using the stressstrain relation, [ ] ! ! ! ! ! ! " # $ $ $ $ $ $ % & + ’ + ’ = E P E P P E P P ( ( ( ) 2 Compatibility is also satisfied because all entries in the strain tensor are constant. Boundary conditions: end faces [ ] T n 1 ± = ! satisfied 33 23 13 ! " ! # $ = = = % % % Boundary conditions: lateral surface [ ] T n sin cos ! ! = ! ! " ! " ! ! " ! " ! sin cos sin sin cos cos 22 12 2 12 11 1 + = # = + = # = P T P T Since the boundary conditions are satisfied, the trial solution is the solution. 1b. The stress state from part (a) is already the principal stress. Therefore, P ! = = = 3 2 1 " " " Using the Von Mises criterion, ( ) ( ) ( ) y y oct P P P P ! ! ! ! ! ! ! ! " = # = = + + = $ + $ + $ = 3 2 3 2 3 1 3 1 2...
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 Spring '09
 Lambros
 Materials Science, Force, Yield surface, Boundary conditions, P1 P2, Von Mises criterion

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