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There are two main ways to write stress/strain tensors as 6 components vectors: Voigt notation, that is the most common; and Mandel-Kelvin notation, that has the advantage of writing stress and strains in the same way, so their rotations are done via the same $6\times 6$ matrices. A reference that I consider good for Voigt's notation is Auld's book (Vol. ...


4

The reason that this particular mesh does not give the correct, uniform displacement solution to this problem is that it is "non-conforming." Specifically, at the intersection of the two cubes in the model, the two element edges that cross that face don't align with each other but instead cross each other. A typical face in a non-conforming mesh ...


4

You already know that at least theoretically, unconstrained matrices have a null space and consequently eigenvalues that are equal to zero. But, in practice, this is a meaningless condition because it can not be checked in an efficient way for large problems. The question you specifically ask is how you can detect whether constraints have been applied, and ...


4

So so many places you have to rewrite. The whole mesh handling (accessing faces and edges from cells, neighbors from cells, ...). Shape functions. Dealing with the question of how the normal vector of a face when seen from one cell matches that on a neighboring cell. You will also likely encounter that 3d problems are always much larger and that solvers, ...


4

Use of Lagrange multipliers produces a saddle-point problem, $$ \begin{pmatrix} A & B^T \\ B & 0 \end{pmatrix} \begin{pmatrix} u \\ \lambda \end{pmatrix} = \begin{pmatrix} b \\ 0 \end{pmatrix} $$ As you've noticed, many preconditioners break down for this sort of system. One can use direct solvers that support pivoting, but if you want iterative ...


3

What you are looking for is a Discrete Kirchhoff Quadrilateral plate or DKQ plate. Seems you are looking for a very straight forward formulation that simply give you the global stiffness matrix. But i'm afraid that most codes I've seen are dealing with integration and transformation. You can search for DKQ source code. There are documents for java which ...


3

TL;DR How can I determine which constrain I need to apply to the system to make problem solved? Or how can I determine which rigid body constrain I should apply to the system? The constraints are given by the boundary conditions of your problem, so you should know them before you have a numerical method as the FEM. In that sense, that is more a physics ...


3

Ignoring subtleties regarding differentiation in two separate tangent spaces, and also avoiding complications arising from non-Euclidean coordinates, we can proceed as follows. The derivative of $\boldsymbol{P}$ with respect to $\boldsymbol{C}$ is $$ \frac{\partial\boldsymbol{P}}{\partial\boldsymbol{C}} = \frac{\partial\boldsymbol{P}}{\partial\...


3

A dyadic product takes as input two vectors and outputs a second order tensor. This is what I know as a dyadic product, and a dyad is the term $\mathbf{a}\mathbf{b}$. A general second order tensor can be written as a linear combination of dyads. Commonly the symbol $\otimes$ is referred as the tensor product and it outputs higher-order tensors. I think that ...


2

The first equality you wrote is not correct, as noted by other users. However, what I think you want to know is why $(I:A)I = (I \otimes I)A$. You can show this just by using dyads properties. $$(B \otimes B) : A =(B_{ij}B_{kl} e_i \otimes e_j \otimes e_k \otimes e_l):(A_{mn} e_m \otimes e_n) = B_{ij}B_{kl} A_{mn} \delta_{km} \delta_{ln} e_i \otimes e_j = B_{...


2

As you said, "If displacement not be constrained, equation above can not be solved, because the system can have rigid body motion" So you should try to apply constraints that will not allow the body to move i.e. translate or rotate. In 2D there are 2 translations (along x and y axis) and one rotation (along z axis) to be killed. In 3D there are 3 ...


2

You take an arbitrary volume $V$ and use the translational and rotational equilibrium equations over it. Then, due to the arbitrariness of the volume the integrals should equal 0 and you get the equation you present and the symmetry for the stress tensor (in classic elasticity). According to @BiswajitBanerjee's comment, the first publications to discuss the ...


1

A different perspective on the question is this: Newton's law says that mass times acceleration equals the sum of all forces. You are interested in the steady state case, so the acceleration is zero and as a consequence, the sum of all forces is zero. This has to hold at each point of the solid if you want the body to not move. The sum of all forces equals ...


1

I believe that the issue you are facing emanates from the type of triangular mesh you are using. This particular discretisation has in-built anisotropy; note the alignment of all of the longest edges is parallel to one of the diagonals of the square. You will observe a different behaviour in the results if you choose the alignment parallel to the other ...


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