Xcellent , superb explanation , total visualization. Thanks a lot Prof.

Professor, u have so much understanding...better explanation....made us easy to understand

Thank you so much Dr. Espinoza. I finally understand the principal, deviatoric, invarianst etc. Struggled with these concepts previously.

Thank you so much professor. Finally, I can understand. Saludos desde Panamá

@dnicolasespinoza5258

2 жыл бұрын

Saludos!

Thank you so much for this lecture!!

Wow!

Thank you so much Dr. Espinoza. Could you tell me in which book or article to find the physical interpretation of the third invariant?

@dnicolasespinoza5258

Жыл бұрын

Great question Esteban! I had a feeling for what it meant but not quite until you asked and I solved the problem! Here it goes: The third invariant reduces to the determinant of the stress tensor, so when written in principal stresses I_3 = sigma_1*sigma_2*sigma_3 Let's call D13 = sigma_1 - sigma_3 and D23 = sigma_2 - sigma_3, and use these D's (Mohr circle diameters) in the equation above. You get I_3 = D13*D23*sigma_3 + (D13+D23)*sigma_3 + sigma_3^3 Hence, I_3 gives you an idea of the difference between sigma_1 and sigma_2 with sigma_3. For a compressional "polyaxial" state with sigma_1 != sigma_2 != sigma_3, I_3 is bound by the perfect triaxial extension case (sigma_1 = sigma_2 > sigma_3) and the perfect triaxial compression case (sigma_1 > sigma_2 = sigma_3) (See lectures on Mohr-Coulomb criterion with triaxial extension and compression), such that Triax ext bound: (sigma_1 - sigma_3)*sigma_3^2 + sigma_3^3 In summary, I_3 tell you how far you are in between triax compression and triax extension! Notice the I_3 = 0 would happen for plane-stress (sigma_3 = 0) or I_3

Waiting for "better than my university professor" comments

Thank you so much for your lecture!