Tag Info

New answers tagged

0

John's answer is pretty good - I will just add my own to give some alternative insights ( I had made a comment to John's answer, but comments tend to get deleted after a while). First - a Bingham plastic acts as a solid under low shear; this means that in the static case (which you explicitly ask for when you ask for "static pressure"), it is a solid. The ...


1

As the Eiffel tower famously shows, a "good" self-supporting structure does not have a uniform section - instead, at every level the size of the supporting surface is large enough to support the weight of the structure above it without reaching a fracture / yield point of the building material. The melting point of ice is a function of pressure - so the ...


1

Below the yield stress your fluid is behaving like an elastic solid. Imagine putting your tank in zero g, so there are no forces, and then removing the base. The result would look like the left hand figure in the diagram below: Now turn on gravity, or apply an external force and the result will be the middle diagram. As long as the stress is below the ...


1

We would have to know more about the ice we want to build the wall with. For example, for ice in icesheets, you have an ice which effectively reaches a plastic region of the stress strain curve at around $0,5 MPa$. I am not a geologist, but I believe that the glaciers can be only thicker than $\sim 50 m$ thanks to it's specific shape and the fact that the ...


0

You probably know that while the Cauchy stresses are objective, its stress rate (material derivative) is not. If $Q(t)$ is an orthogonal tensor representing a change of frame, the stress is the new frame is $$ T^* = Q T Q^T $$ However, if you take material derivatives on both sides, you have, $$ \dot{T^*} = \dot{Q} T Q^T + Q \dot{T} Q^T + Q T \dot{Q}^T ...



Top 50 recent answers are included