Timeline for What is the actual stress in a certain point in a body?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 26, 2019 at 21:30 | comment | added | Biswajit Banerjee | If the "allowed" stress is computed using the same formula, then yes. Often, people just consider the deviatoric part of the stress (what you get after removing the part of stress that only causes volume change) and then equate the allowable stress to the magnitude of the deviatoric stress. | |
| Oct 26, 2019 at 19:28 | comment | added | user1477107 | So is that stress the one we compare with the allowed stress at that point? | |
| Oct 25, 2019 at 21:08 | comment | added | Biswajit Banerjee | There's something called the "equivalent stress" or "von Mises stress" which is analogous to the magnitude of a vector. But the correspondence is not exact. The magnitude of a vector $\mathbf{v}$ is $\sqrt{\mathbf{v}\cdot\mathbf{v}}$. The analogous magnitude of a tensor $\boldsymbol{\sigma}$ is $\sqrt{\boldsymbol{\sigma}:\boldsymbol{\sigma}}$. If you think of the tensor as a nine-dimensional vector, the analogy is exact. | |
| Oct 25, 2019 at 16:30 | comment | added | user1477107 | I appreciate the effort you put into the answer, thank you very much. If I understand correctly, just like a vector can be described by 2 components, stress can be described with a tensor (6 independent components). But if we know the angle between the vector components and their magnitudes, we can calculate the magnitude of the vector (the "real" value). Can the same be done for stress? Can I calculate the "real" stress if I'm given a tensor? | |
| Oct 25, 2019 at 12:17 | history | edited | alephzero | CC BY-SA 4.0 |
added 816 characters in body
|
| Oct 25, 2019 at 12:02 | history | answered | alephzero | CC BY-SA 4.0 |