2.2 – MATERIAL MODELS Using the information below, choose the correct answer to the closest 3 significant figures. For questions 1 through 5, assume plane stress. For questions 6 through 10, assume plane strain. Name Email 1. Assume plane stress: The principal strain one is373 µ587 µ817 µ800 µNone of the above2. Assume plane stress: The principal strain three is0-66.7 µ-360 µ360 µNone of the above3. Assume plane stress: The factor of safety to avoid yielding using maximum shear stress theory is7.52.343.411.39None of the above4. Assume plane stress: The factor of safety to avoid yielding using maximum octahedral shear stress theory is2.641.592.341.39None of the above5. Assume plane stress: The critical crack length in Mode I is0.995 inch0.470 inch0.497 inch0.235 inchNone of the above6. Assume plane strain: The principal stress three is2 ksi (T)10.8 ksi (T)10.8 ksi (C)2 ksi (C)None of the above7. Assume plane strain The principal strain one is373 µ587 µ817 µ800 µNone of the above8. Assume plain strain: The factor of safety to avoid yielding using maximum shear stress theory is7.52.343.411.39None of the above9. Assume plane strain: The von-Mises stress is20.79 ksi36.81 ksi22.68 ksi37.89 ksiNone of the above10. Assume plane strain: The factor of safety to avoid yielding using maximum octahedral shear stress theory is1.852.652.891.60None of the aboveDo you have any comments or suggestions? Please send them to author@madhuvable.org Time is Up!Time's up