1
$\begingroup$

I understand plane strain/stress relations are typically utilized for 2D materials. I was wondering if we consider a 2D material acting in a 2D space, would we have a stiffness matrix such as \begin{equation*} \left[ \begin{array}{ccc} C_{1111} & C_{1122} & C_{1112} \\ & C_{2222} & C_{2212} \\ & & C_{1212} \end{array} \right] \end{equation*} rather than just plane strain/stress relations imposed on the classical stiffness tensor in 3D. Are there any situations where such a formulation would be appropriate in the real world? Thanks for the insight!

$\endgroup$
1
  • $\begingroup$ I'm really grasping to understand what you are asking. What is a "2D" material? Are you envisioning a "2D" space that isn't a plane? $\endgroup$
    – agentp
    Sep 13 '17 at 20:50
1
$\begingroup$

As mentioned by Apostolos above, the real world is complicated. The stiffness tensor that you wrote correct but for most applications assuming plane stress or strain is enough, especially when dealing with isotropic materials. If you don't make assumptions, you have to find the 6 constants, which is not necessarily easy.

If you extend the stiffness tensor to 3D you get the following

$$ \begin{bmatrix} \sigma_{1} \\ \sigma_{2} \\ \sigma_{3} \\ \sigma_{4} \\ \sigma_{5} \\ \sigma_{6} \end{bmatrix} = \begin{bmatrix} C_{11} & C_{12} & C_{13} & C_{14} & C_{15} & C_{16} \\ C_{21} & C_{22} & C_{23} & C_{24} & C_{25} & C_{26} \\ C_{31} & C_{32} & C_{33} & C_{34} & C_{35} & C_{36} \\ C_{41} & C_{42} & C_{43} & C_{44} & C_{45} & C_{46} \\ C_{51} & C_{52} & C_{53} & C_{54} & C_{55} & C_{56} \\ C_{61} & C_{62} & C_{63} & C_{64} & C_{65} & C_{66} \end{bmatrix} \begin{bmatrix} \varepsilon_{1} \\ \varepsilon_{2} \\ \varepsilon_{3} \\ \varepsilon_{4} \\ \varepsilon_{5} \\ \varepsilon_{6} \end{bmatrix} $$

Here it's much worse than the 2D case because you have to find 36 constants, assuming that the matrix is not symmetric. This is the most complete characterization of a material. The problem is that that's a lot of testing which is very hard! Therefore assumptions are made in order to simplify and reduce the number of constants required.

In most real world applications, the worse case scenario is assuming the material is orthotropic. The stiffness matrix for an orthotropic material is simpler than that for fully anisotropic so once inverted and having introduced $E$, $\nu$, and $G$ the result is $$ \begin{bmatrix} \varepsilon_{1} \\ \varepsilon_{2} \\ \varepsilon_{3} \\ \varepsilon_{4} \\ \varepsilon_{5} \\ \varepsilon_{6} \end{bmatrix} = \begin{bmatrix} \frac{1}{E_1} & -\frac{\nu_{21}}{E_2} & -\frac{\nu_{31}}{E_3} & 0 & 0 & 0 \\ -\frac{\nu_{12}}{E_1} & \frac{1}{E_2} & -\frac{\nu_{32}}{E_3} & 0 & 0 & 0 \\ -\frac{\nu_{13}}{E_1} & -\frac{\nu_{23}}{E_2} & \frac{1}{E_3} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2G_{23}} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2G_{13}} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G_{12}} \end{bmatrix} \begin{bmatrix} \sigma_{1} \\ \sigma_{2} \\ \sigma_{3} \\ \sigma_{4} \\ \sigma_{5} \\ \sigma_{6} \end{bmatrix} $$ This means you have to measure "only" nine material constants, assuming the matrix is symmetrical, making life a whole load easier. Examples of situations where the material is assumed to be orthotropic are 3D printing and composites.

$\endgroup$
0
$\begingroup$

In the real world this is the "real" stiffness because nothing is perfect and homogeneous. We just simplify things assuming plane stress or stain because we don't have the whole picture or it's not easy to get it.

A simple example for you to think about is wood. Wood has different stiffness in each direction (anisotropic material).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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