- #1
Peeter
- 305
- 3
For fluid with viscosity [itex]\mu[/itex] our stress strain relationship takes the form
[tex]\sigma_{ij} = -p \delta_{ij} + 2 \mu u_{ij}.[/tex]
I was wondering how to express this in cylindrical coordinates. The strain tensor I can calculate in cylindrical coordinates (what I get matches eq 1.8 in [1]). But how would the [itex]\delta_{ij}[/itex] portion of the stress strain relationship be expressed in cylindrical coordinates?
For example, if we considered a non-viscous fluid, the very simplest stress tensor, we have in rectangular coordinates
[tex]\sigma_{ij} = -p \delta_{ij}.[/tex]
It's not obvious to me how this would be expressed in cylindrical form. I wanted to try some calculations with the traction vector [itex]T_i = \sigma_{ij} n_j[/itex] in a cylindrical coordinate system, but I'm not sure how to express it. I figured the place to start was with the stress tensor.
References:
[1] L.D. Landau, EM Lifgarbagez, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.
[tex]\sigma_{ij} = -p \delta_{ij} + 2 \mu u_{ij}.[/tex]
I was wondering how to express this in cylindrical coordinates. The strain tensor I can calculate in cylindrical coordinates (what I get matches eq 1.8 in [1]). But how would the [itex]\delta_{ij}[/itex] portion of the stress strain relationship be expressed in cylindrical coordinates?
For example, if we considered a non-viscous fluid, the very simplest stress tensor, we have in rectangular coordinates
[tex]\sigma_{ij} = -p \delta_{ij}.[/tex]
It's not obvious to me how this would be expressed in cylindrical form. I wanted to try some calculations with the traction vector [itex]T_i = \sigma_{ij} n_j[/itex] in a cylindrical coordinate system, but I'm not sure how to express it. I figured the place to start was with the stress tensor.
References:
[1] L.D. Landau, EM Lifgarbagez, JB Sykes, WH Reid, and E.H. Dill. Theory of elasticity: Vol. 7 of course of theoretical physics. 1960.