figura

nafems4.md

@@ -782,7 +782,7 @@ For a fixed number of nodes, first-order grids have far more elements than secon
 
 The linear FEM problem leads of course of a system of\ $NG$ linear equations, cast in matrix form by the stiffness matrix\ $K$ and a right-hand size vector\ $\vec{b}$ containing the loads (both volumetric and the ones at the surfaces from the boundary conditions):
 
-$$K \cdot \vec{u} = \vec{b}$$
+$$K \cdot \vec{u} = \vec{b}$${#eq:kub}
 
 The objective of the solver is to find the vector\ $u$ of nodal displacements that satisfy the momentum equilibrium. Luckily (well, not purely by chance but by design) the stiffness matrix is almost empty. It is called a [sparse matrix](https://en.wikipedia.org/wiki/Sparse_matrix) because most of its elements are zero. If it was fully filled, then a problem with just 100k nodes would need more than 700\ Gb of RAM just to store the matrix elements, rendering FEM as practically impossible. And even though the stiffness matrix is sparse, its inverse is not so we cannot solve the elastic problem by “inverting” the matrix. Particular methods to represent and more importantly to solve linear systems involving these kind of matrices have been developed, which are the methods used by finite-element (and the other finite-cousins) programs. These methods are [iterative](https://en.wikipedia.org/wiki/Iterative_method) in nature, meaning that at each step of the iteration the numerical answer improves, i.e. $\left|K \cdot \vec{u} - \vec{b}\right| \rightarrow 0$. Even though in particular for [Krylov-subspace](https://en.wikipedia.org/wiki/Krylov_subspace) methods, $K \cdot \vec{u} - \vec{b} = \vec{0}$ after\ $NG$ steps, a few dozen of iterations are usually enough to assume that the problem is effectively solve (i.e. the residual is less than a certain threshold).
 
@@ -969,7 +969,7 @@ Finally we attempt to “break” the pipes successively solving many steady-sta
  9. The stress linearisation has to be performed individually for each principal stress\ $\sigma_1$, $\sigma_2$ and $\sigma_3$ to fulfill the requirements ASME\ III\ NB-3126 (see [@sec:in-air] below).  
  10. This “break” step is linear.
 
-[Surely you’re joking, Mr.\ Theler!](https://en.wikipedia.org/wiki/Surely_You're_Joking%2C_Mr._Feynman!) How can this extremely complex problem be linear? Well, let us see. First, there are two main kinds of non-linearities in FEM:
+[Surely you’re joking, Mr.\ Theler!](https://en.wikipedia.org/wiki/Surely_You're_Joking%2C_Mr._Feynman!) Linear problems are simple and almost useless. How can this extremely complex problem be linear? Well, let us see. First, there are two main kinds of non-linearities in FEM:
 
  1. Geometrical non-linearities
  2. Material non-linearities
@@ -980,12 +980,22 @@ About material non-linearities, on the one hand we have the temperature-dependen
 
 $$ \sigma = E(T) \cdot \epsilon $$
 
-What changes with temperature is the slope of\ $\sigma$ with respect to\ $\epsilon$ (think and imagine!). On the other hand, we have a given non-trivial temperature distribution\ $T(\vec{x}, t)$ within the pipes that is a snapshot of a transient heat conduction problem at a certain time\ $t$ (think and picture yourself taking photos of the temperature distribution changing in time). Let us now forget about the time, as after all we are solving a steady-state elastic problem.
+What changes with temperature is the slope of\ $\sigma$ with respect to\ $\epsilon$ (think and imagine!). On the other hand, we have a given non-trivial temperature distribution\ $T(\vec{x}, t)$ within the pipes that is a snapshot of a transient heat conduction problem at a certain time\ $t$ (think and picture yourself taking photos of the temperature distribution changing in time). Let us now forget about the time, as after all we are solving a steady-state elastic problem. Now you can trust me or ask a FEM teacher, but the continuous displacement formulation can be loosely written as
 
 $$ K\big[E\left(T(\vec{x})\right), \vec{x}\big] \cdot \vec{u}(\vec{x}) = \vec{b}(\vec{x})$$
 
+If you notice, the complex dependence of the stiffnes matrix\ $K$ can be reduced to just the spatial vector\ $\vec{x}$:
+
+$$ K(\vec{x}) \cdot \vec{u}(\vec{x}) = \vec{b}(\vec{x})$$
+
+And this last equation is linear in\ $\vec{u}$. In effect, the discretization step means to integrate over\ $\vec{x}$. As\ $K$, $\vec{u}$ and\ $\vec{b}$ depend only on\ $\vec{x}$, then after integration one gets just numbers with the matrix representation of\ [@eq:kub]. Again, you can either trust me, ask a teacher or go through with the maths.
+
+**33410 1D-403**
+
+
 # Fatigue
 
+
 ## In air {#sec:in-air}
 
 ## In water