cosas

nafems4.md

@@ -34,7 +34,7 @@ After further years passed by, engineers (probably the same people that forked s
 
 # Solid mechanics, or what we are taught at college
 
-So, let us start our journey. Our starting place: undergraduate solid mechanics courses. Our first step: Newton’s laws of motion. For each of them, all we need to recall here is that
+So, let us start our journey. Our starting place: undergraduate solid mechanics courses. Our goal: to obtain the internal state of a solid subject to a set of movement restrictions and loads (i.e. to solve the solid mechanics problem). Our first step: Newton’s laws of motion. For each of them, all we need to recall here is that
 
  1. a solid is in equilibrium if it is not moving in at least one inertial coordinate system,
  2. in order for a solid not to move, the sum of all the forces ought to be equal to zero, and
@@ -44,7 +44,11 @@ We have to accept that there is certain intellectual beauty when complex stuff c
 
 ## The stress tensor
 
-In any case, what we should understand (and imagine) is that external forces lead to internal stresses. And in any three-dimensional body subject to such external loads, the best way to represent internal stresses is through a $3 \times 3$ _stress tensor_. This is the first point in which we should not fear math. Trust me, it will pay back later on. Let me introduce then the three-dimensional stress tensor:
+In any case, what we should understand (and imagine) is that external forces lead to internal stresses. And in any three-dimensional body subject to such external loads, the best way to represent internal stresses is through a $3 \times 3$ _stress tensor_. This is the first point in which we should not fear math. Trust me, it will pay back later on.
+
+Does the term _tensor_ scare you? It should not. A tensor is just a slightly more complex vector, and I assume you are not afraid of vectors. If you recall, a vector somehow generalises the idea of a scalar in the following sense: a given vector $\vec{v}$ can be projected into any direction $\vec{n}$ to obtain a scalar $p$. We call this scalar $p$ the “projection” of the vector $\vec{v}$ in the direction $\vec{n}$. Well, a tensor can be also projected into any direction $\vec{n}$. The difference is that instead of a scalar, a vector is now obtained.
+
+Let me introduce then the three-dimensional stress tensor:
 
 $$
 \begin{bmatrix}
@@ -54,10 +58,10 @@ $$
 \end{bmatrix}
 $$
 
-It looks like a regular $3 \times 3$ matrix, doesn’t it? Some brief comments about it:
+It looks (and works) like a regular $3 \times 3$ matrix. Some brief comments about it:
 
  * The $\sigma$s are normal stresses, i.e. they try to stretch or tighten the material.
- * The $\tau$s are shear stresses, i.e. they try to bend and rotate the material.
+ * The $\tau$s are shear stresses, i.e. they try to twist the material.
  * It is symmetric (so there are only six independent elements) because
     - $\tau_{xy} = \tau_{yx}$,
     - $\tau_{yz} = \tau_{zy}$, and
@@ -65,36 +69,11 @@ It looks like a regular $3 \times 3$ matrix, doesn’t it? Some brief comments a
  * The elements of the tensor depends on the orientation of the coordinate system.
  * There exists a particular coordinate system in which the stress tensor is diagonal, i.e. all the shear stresses are zero. In this case, the three diagonal elements are called the principal stresses.
 
-Still scared of the term _tensor_? Do not be! It is just a slightly more complex vector, and I assume you are not afraid of vectors. If you recall, a vector somehow generalises the idea of a scalar in the following sense: a given vector $\vec{v}$ can be projected into any direction $\vec{n}$ to obtain a scalar $p$. We call this scalar $p$ the “projection” of the vector $\vec{v}$ in the direction $\vec{n}$. Well, a tensor can be also projected into any direction $\vec{n}$. The difference is that instead of a scalar, a vector is now obtained.
-
-What does this all have to do with mechanical engineering? Well, once we know what the stress tensor is for every point of a part, in order to obtain the internal forces per unit area acting in a plane passing through that point and with a normal given by the direction $\vec{n}$, all we have to do is “project” the stress tensor through $\vec{n}$. That is all we need to know about the stress tensor for now. 
-
-dnl Oh yes, and the fact that tensors follow the rules of matrix multiplication. 
-
-dnl That is to say, the internal stress $\vec{T}$ (i.e. force per unit area) is
-dnl 
-dnl $$
-dnl \begin{bmatrix}
-dnl T_x \\
-dnl T_y \\
-dnl T_z \\
-dnl \end{bmatrix}
-dnl =
-dnl \begin{bmatrix}
-dnl \sigma_x  & \tau_{xy}  & \tau_{xz} \\
-dnl \tau_{yx} & \sigma_{y} & \tau_{yz} \\
-dnl \tau_{zx} & \tau_{zy}  & \sigma_{z} \\
-dnl \end{bmatrix}
-dnl \cdot
-dnl \begin{bmatrix}
-dnl n_x \\
-dnl n_y \\
-dnl n_z \\
-dnl \end{bmatrix}
-dnl $$
-dnl  
-dnl And yes, tensors follow the rule of matrix multiplication. So if you take a look, in general the internal forces do not 
-dnl
+
+What does this all have to do with mechanical engineering? Well, once we know what the stress tensor is for every point of a solid, in order to obtain the internal forces per unit area acting in a plane passing through that point and with a normal given by the direction $\vec{n}$, all we have to do is “project” the stress tensor through $\vec{n}$. In plain simple words:
+
+ * If you can compute the stress tensor at each point of our geometry, then congratulations: you have solved the solid mechanics problem.
+
 
 ## An infinitely-long pressurised pipe