I do not have any engineering background but I am studying some mathematics to try and understand and solve this problem. I have a mesh (M1) in cylindrical coordinates and this deforms into a mesh (M2) (again in cylindrical coordinates). I need to compute strain in radial, longitudinal and circumferential directions.

Basically I have done the following computations on each point of the mesh. The deformation gradient F is computed:
$$F = x·X^T·(X·X^T)^{-1}$$ where:

  • $x$: the deformed cylindrical coordinates (3x1 matrix)
  • $X$: the undeformed cylindrical reference coordinates (3x1 matrix)
  • $X^T$: transpose of $X$
  • $X^{-T}$ : inverse of transpose of $X$

The Langrangian finite strain tensor $E$ is then computed:
$$E = \dfrac{1}{2}(F^T·F-I)$$


  • $F^T$ : transpose of $F$
  • $I$ : identity matrix (3x3)

I then take the diagonal of $E$ and this gives me the principal strains in the orthogonal cylindrical coordinates.

Questions: Am I doing it right? Do I need to do some further computations on the diagonal of $E$ to get the principal strains at each point of this mesh that is deforming?


1 Answer 1


I'm not sure what the interpretation of your deformation gradient is. To my knowledge the deformation gradient gives the local deformation, i.e. the deformation around a certain point. To obtain this deformation, local information (i.e. information in one point only) is not enough, but you also require some information on the surroundings. The formula I know is $$ F_{ij}=\frac{\partial x_i}{\partial X_j} $$ with $F_{ij}$ being the deformation gradient, and $x_i$ and $X_j$ are local coordinates in the deformed and undeformed configuration, respectively. Thus to calculate the continuous $F_{ij}$, you need the differentials of the local coordinates $\partial x_i$ and $\partial X_j$. I'm no expert on discrete meshes, but one possibility is to calculate $F_{ij}$ for each element in your mesh by using finite differences $\Delta x_i$ and $\Delta X_j$. If unsure you should probably ask again more precisely.

Your finite strain tensor calculation seems correct to me. Note that when the deformation gradient is calculated for an element, the finite strain tensor is also for that element. Again I'm no expert on the interpretation of the results. The wikipedia article on finite strain theory seems to have some answers for you.

When calculating in curvilinear coordinate systems, things usually become a bit more complicated than in cartesian coordinates. However, since cylindrical coordinates are locally cartesian, your calculation is fine. For more complex curvilinear coordinate systems you would need to evaluate your equations using co- and contravariant bases.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.