linear

nafems4.md

@@ -17,7 +17,7 @@ Again, take all this information as coming from a fellow that has already taken
 There are some useful tricks that come handy when trying to solve a mechanical problem. Throughout this text, I will try to tell you some of them.
 
 
-One of the most important ones is using your _imagination_. You will need a lot of imagination to “see“ what it is actually going on when analysing an engineering problem. How the loads “press” one element with the other, how the material reacts depending on its properties, how the nodal displacements generate stresses (both normal and shear), how results converge, etc. And what these results actually mean besides the pretty-coloured figures.^[A former boss once told me “I need the CFD” when I handed in some results. I replied that I did not do computational fluid-dynamics but computed the neutron flux kinetics within a nuclear reactor core. He joked “I know, what I need are the _Colors For Directors_, those pretty coloured figures along with your actual results.”]
+One of the most important ones is using your _imagination_. You will need a lot of imagination to “see“ what it is actually going on when analysing an engineering problem. This skill comes from my background in nuclear engineering where I had not choice but to imagine a [positron-electron annihilation](https://en.wikipedia.org/wiki/Electron%E2%80%93positron_annihilation) or an [Spontaneous fission](https://en.wikipedia.org/wiki/Spontaneous_fission). But in mechanical engineering, it is likewise important to be able to imagine how the loads “press” one element with the other, how the material reacts depending on its properties, how the nodal displacements generate stresses (both normal and shear), how results converge, etc. And what these results actually mean besides the pretty-coloured figures.^[A former boss once told me “I need the CFD” when I handed in some results. I replied that I did not do computational fluid-dynamics but computed the neutron flux kinetics within a nuclear reactor core. He joked “I know, what I need are the _Colors For Directors_, those pretty coloured figures along with your actual results.”]
 This journey will definitely need your imagination. We will see  equations, numbers, plots, schematics, 3D geometries, interactive 3D views, etc. Still, when the theory says “thermal expansion produces linear stresses” you have to picture in your head three little arrows pulling away from the same point in three directions, or whatever mental picture you have about what you understand are thermally-induced stresses. What comes to your mind when someone says that out of the nine elements of the stress tensors there are only six that are independent? Whatever it is, try to practice that kind of graphical thoughts with every concept.
 
 Another heads up is that we will dig into some math. Probably it would be be simple and you would deal with it very easily. But probably you do not like equations. No problem! Just ignore them for now. Read the text skipping them, it should work. It is fine to ignore math (for now). But, eventually, a time will come in which it cannot (or should not) be avoided. Here comes another experience tip: do not fear mathematics. Even more, keep exercising. You have used differences of squares in high school. You know (or at least knew) how to integrate by parts. Remember what Laplace transforms are used for? Once in a while, perform a division of polynomials using [Ruffini’s rule](https://en.wikipedia.org/wiki/Ruffini's_rule). Or compute the second derivative of the quotient of two functions. Whatever. It should be like doing crosswords on the newspaper. Grab those old physics college books and read the exercises at the end of each chapter. It will pay off later on.
@@ -476,21 +476,63 @@ The equivalent accelerations for the piping section of [@fig:modes] for the spec
 The ASME code says that these accelerations (depicted in [@fig:acceleration]) are to be applied twice. Once with the original sign and once with all the elements with the opposite sign during two seconds of the transient each time.
 
 
-## Linearity of displacements and stresses
+## Linearity (not yet linearization)
 
-Even though we did not yet discuss it in detail, we want to solve an elastic problem subject to an internal pressure condition, with a non-uniform temperature distribution that lead to both thermal stresses and variations in the mechanical properties of the materials. And as if this was not enough, we want to add at some instants a statically-equivalent distributed load that comes from a design earthquake.
+Even though we did not yet discuss it in detail, we want to solve an elastic problem subject to an internal pressure condition, with a non-uniform temperature distribution that leads to both thermal stresses and variations in the mechanical properties of the materials. And as if this was not enough, we want to add at some instants a statically-equivalent distributed load that comes from a design earthquake. This last point means that at the transient instant where the stresses (from the fatigue’s point of view) are maximum we have to add the distributed loads that we computed from the seismic spectra to the other thermal and pressure loads. But we have a linear elastic problem (well, we still do not have it but we will in\ [@sec:break]), so we might be tempted exploit the problem’s linearity and compute all the effects separately and them sum them up to obtain the whole combination. We may thus compute just the stresses due to the seismic loads and then add them up to the stresses of any instant of the transient, in particular to the one with the highest ones. After all, in linear problems the result of the sum of two cases is the results of the sum of the cases, right? Wrong.
 
-Just for the sake of it, let us jump out of our nuclear piping problem and step back into the general finite-element theory ground (remember we were going to jump back and forth). We have a linear elastic problem (well, we still do not have it but we will in\ [@sec:break]), so we might exploit the problem’s linearity and compute all the effects separately and them sum them up to obtain the whole combination, right? Well, not so much. Just like most of the time when we want to weight two masses we can sum their weights individually to obtain the same value we might get when putting both of them into a scale, most of the time we might just sum partial finite-element results. If we individually weighted two protons and two neutrons on the one hand and one $\alpha$ particle (i.e. a ⁴He nucleus) we would definitely not get the same result.
+Let us jump out of our nuclear piping problem and step back into the general finite-element theory ground for a moment (remember we were going to jump back and forth). Assume you want to know how much your dog weights. One thing you can do is weight yourself (let us say you weight 81.2\ kg), then grab your dog and weight yourself and your dog (let us say you and your dog weight 87.3\ kg). Do you swear your dog weights 6.1\ kg plus/minus the scale’s uncertainty? I can tell you that the weight of two individual protons and two individual neutrons in not the same as the weight of an\ [$alpha$ particle](Alpha_particle). Will not there be a master-pet interaction that renders the weighting problem non-linear?
 
-divert(-1)
-Let us both (i.e. you and me) make an experiment. Grab a FEM program of your choice, get a square-sectioned beam of any size and length, fix one of the ends and put an uniform vertical load in the top surface. In my case, I have a square section of 1mm\ $\times$\ 1mm and a cantilever length of 10mm. A vertical load of 10N uniformly distributed in the top surface gives a maximum displacement of  and a maximum Von Mises stress of  
+Let us both (i.e. you and me) make an experiment. Grab a FEM program of your choice (mine is [CAEplex](https://caeplex.com)) and load a 1mm $\times$ 1mm $\times$ 1mm cube. Set any values for the Young Modulus and Poisson ratio as you want. I chose\ $E=200$MPa and\ $\nu=0.28$. Restrict the three faces pointing to the negative axes to their planes, i.e.
+
+ * in face “left” ($x<0$), set null displacement in the $x$ direction ($u=0$),
+ * in face “front” ($y<0$), set null displacement in the $y$ direction ($v=0$),
+ * in face “bottom” ($z<0$), set null displacement in the $z$ direction ($w=0$).
+
+Now we are going to create and compare three load cases:
+
+ A. Pure normal loads (<https://caeplex.com/p?d8f>)
+ B. Pure shear loads (<https://caeplex.com/p?b494>)
+ C. The combination of A & B (<https://caeplex.com/p?989>)
+
+The loads in each cases are applied to the three remaining faces, namely “right”, “back” and “top,” and their magnitude in Newtons are:
+
+\begin{tabular}{l|ccc|ccc|ccc}
+&
+\multicolumn{3}{c|}{face “right” ($x>0$)} &
+\multicolumn{3}{c|}{face “back” ($y>0$)} &
+\multicolumn{3}{c}{face “top” ($z>0$)} \\
+
+&
+$F_x$ &
+$F_y$ &
+$F_z$ &
+$F_x$ &
+$F_y$ &
+$F_z$ &
+$F_x$ &
+$F_y$ &
+$F_z$ \\
+
+\hline
+
+Case A, pure normal & +10 &  0  &   0 &    0 & +20 &  0   &   0 &   0 & +30 \\
+Case B, pure shear  &   0 & +15 & -15 &  +25 &   0 & -5   & -15 & +25 & +30 \\
+Case C, combination & +10 & +15 & -15 &  +25 & +20 & -5   & -15 & +25 & +30 \\
+\end{tabular}
+
+
+In the first case, the principal stresses are uniform and equal to the three normal loads. As the forces are in Newton and the area of each face of the cube is 1mm^2, the usual sorting leads to
+
+\begin{align*}
+\sigma_{1A} &= 30 \text{MPa} \\
+\sigma_{2A} &= 20 \text{MPa} \\
+\sigma_{3A} &= 10 \text{MPa} \\
+\end{align*}
+
+In the second case, the principal stresses are not uniform and have a non-trivial distribution. The results obtained in CAeplx
 
-https://caeplex.com/p?9c1
 
-https://caeplex.com/r?dd3e38
 
-no era que todo era lineal y podemos sumar todo?
-si y no
 
 podemos sumar los desplazamientos y los tensores de tensiones, pero ojo que las tensiones principales no son lineales a la suma (si a la multiplicacion)
 
@@ -523,12 +565,12 @@ ans =
    0.782990
    5.767255
 
-octave:36> 
+octave:36>
 
 divert(0)
 
 
-
+I did not know this in college! I learned it the hard way.
 
 cantilever beam, principal stresses, linearity of von mises