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Unformatted text preview: AE 321 – Homework 8 solution 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 in-plane hydrostatic case, and that all other entries are zero because the cylinder is traction-free in the x 3-direction. It can be easily shown that equilibrium is satisfied since P is constant. Using the stress-strain 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 2 2 1 3 2 3 2 2 2 1 max 1c. When the pressure P were also applied on the end faces, then the stress state...
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This note was uploaded on 03/07/2011 for the course AE 311 taught by Professor Dutton,j during the Spring '08 term at University of Illinois, Urbana Champaign.
- Spring '08