Questions and discussion for this lecture live here. Fire away by hitting Reply below 
Are these equations only for materials that are homogenous? Would this apply to something that is anisotropic?
Yes, homogeneous isotropic material is assumed.
S
Is it a bit of a stretch to apply this to say carbon fiber polymers, if I knew the layup (or directions of fibers)?
I have to be honest I don’t have any experience with this material. The analysis methods in this course will give you a first approximation of how forces are distributed through your structure, provided it can be approximated as a pin-jointed structure.
But how the material responds to the forces that are developed is not something I could advise on - my hunch would be that for low stress levels, the approximation would be reasonable but as you start to push the material further, you may introduce additional behaviours not captured within our isotropic homogeneous material assumption.
I guess, you can boil the question down to…
How closely does your CFP material agree with the behaviour of a steel bar in uniaxial tension and compression?
If the agreement is good - by extension, the analysis procedure from the course may be a good approximation.
I like how you explained this. It gives me a better understanding of the analysis limits at these beginning stages of learning.
This is my first dive at FEA, I just completed ‘Structural Analysis’ at the undergrade level. We reached up to Moment Distribution in class.
For some context, I am also in an undergraduate research program where I am making a 2 ft CFRP (Carbon Fiber Reinforced Polymer) Beam and subjecting it to 3 point loading.
My professor/mentor has explained to me that currently there is no complete set of failure theories for CFRP’s because of their engineered nature.
However, we clearly model this material’s behavior with FEA at the aerospace level. Therefore, after completing your courses with FEA/Python, I’ll be trying to understand how to apply this to CFRP.
Sounds like fascinating work - material science is one big area where there are still many gains to be made for Civil and Structural engineers. Good luck with your work and don’t hesitate to get in touch with any further questions 
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Hi Sean, great lecture. I understand that plane stress and plane strain are different conditions. However, I’m curious to understand why the matrix [C] takes on a different form when you make the further simplifying assumption to neglect strain in z-dir for plane stress & stress in z-dir for plane strain. Ultimately you have {Sxx Syy Sxy} = [C].{exx eyy exy} in both cases but [C] is different.
In the case of plane stress you simplified the [D] matrix then inverted it to [C], and in the case of plane strain you directly simplified [C]. In doing so you ended up with different [C] matrices but I didn’t understand why it was done this way.
Any help understanding this would be appreciated. Thanks in advance.
Hi @Richard_Farnell
Thanks for the question - if I understand it, you’re asking…
- Why are the C matrices different (plane stress versus plane strain)
- Why was the derivation of the C matrix procedurally different in each case.
To tackle 1; the C-matrices are different simply because we’re modelling two different structural forms. Remember, the ‘full’ 3D matrix (given in lecture 7) - we can think of this as the ‘ground-truth’ or accurate matrix. The plane stress and plane strain versions that were derived after, are simplifications of this, to address two very different structural forms, plane stress, often a suitable simplification to apply to plates or slabs while plane strain is often suitable for modelling the in-plane behaviour for structures with a large out of plane dimension (e.g. tunnels). So, it shouldn’t be surprising to us that we have two different C matrices.
As for 2; the most direct way to reduce the material matrix, is by making sure that the matrix is being multiplied by the vector containing the stress/strain terms you’re trying to eliminate. This has the effect of directly multiplying the terms of the material matrix by zero - reducing the size of the matrix.
So, for example, in our case, we start with a full 3D matrix.
[strain] =[D][stress]
So, let’s reduce from this by eliminating the irrelevant stresses - the terms on the right. The stress vector now contains, zeros, which are multiplied by terms on the D matrix, directly eliminating them.
Then, we invert to get the C matrix…
[stress] = [C][strain] -> plane stress
For, plane strain, we start with the full matrix, this time…
[stress] = [C][strain]
This with, we want to eliminate strain terms (again, on the right side), so we directly reduce the matrix to get the plane strain C matrix.
I hope that helps clear up any confusion - follow up if not.
Seán