gtheler
/
nafems4
Прати 1
Волим 0
Креирај огранак 0
Код Дискусије 0 Захтеви за спајање 0 Издања 0 Вики Activity
Преглед изворни кода

cosas

tags/CANDIDATE
gtheler пре 7 година
родитељ
23b17b234b
комит
3ae2a77d85
6 измењених фајлова са 856 додато и 28 уклоњено
Подељен поглед Покажи статистику Diff
  1. +21
    -0
      SN.dat
  2. BIN
      SN.pdf
  3. +27
    -0
      SN.ppl
  4. +732
    -0
      SN.svg
  5. BIN
      cad1.png
  6. +76
    -28
      nafems4.md

+ 21
- 0
SN.dat Прегледај датотеку

@@ -0,0 +1,21 @@
10 3999 6000
20 2827 4300
50 1896 2748
100 1413 1978
200 1069 1440
500 724 974
1000 572 745
2000 441 590
5000 331 450
10000 262 368
20000 214 300
50000 159 235
100000 138 196
200000 114 168
500000 93 142
1e+06 86 126
1e+07 77 99
1e+08 68 97.1
1e+09 61 95.8
1e+10 54 94.4
1e+11 48 93.7

BIN
SN.pdf Прегледај датотеку


+ 27
- 0
SN.ppl Прегледај датотеку

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set preamble "\usepackage{amsmath}"

set width 14*unit(cm)

set axis x arrow nomirrored
set axis y arrow nomirrored
set key bottom left

#set xrange [0:1]
#set yrange [1:1e20]

set grid

set terminal pdf

set output "SN.pdf"
set key top right
set xlabel "Number of cycles $N$"
set ylabel "Stress amplitude $S_a$ [MPa]"
set logscale x 10
set logscale y 10
set yrange [10:1e4]
set xrange [10:1e11]

plot "SN.dat" u 1:2 w lp pt 2 lw 2 lt 2 color black ti "SA~106~GrB ($E=207$~MPa)",\
"SN.dat" u 1:3 w lp pt 3 lw 3 lt 3 color blue ti "SA~312~TP~304L, ($E=195$~MPa)"


+ 732
- 0
SN.svg Прегледај датотеку

@@ -0,0 +1,732 @@
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@@ -2,7 +2,7 @@

First of all, please take this text as a written chat between you an me, i.e. an average engineer that have already taken the journey from college to performing actual engineering using finite element analysis and has something to say about it. Picture yourself in a coffee bar, talking and discussing concepts and ideas with me. Maybe needing to go to a blackboard (or notepad?). Even using a tablet to illustrate some three-dimensional results. But always as a chat between colleagues.

Please also note that I am not a mechanical engineer, although I shared many undergraduate courses with some of them. I am a nuclear engineer with a strong background on mathematics and computer programming. I went to college between 2002 and 2008. Probably a lot of things have changed since then---at least that is what these millenials guys and girls seem to be boasting about---but chances are we all studied solid mechanics and heat transfer with a teacher using a piece of chalk on a blackboard and students writing down notes with pencils on paper sheets. And there is really not much that one can do with pencil and paper regarding mechanical analysis. Any actual case worth the time of an engineer need to be more complex than an ideal canonical case with analytical solution.
Please also note that I am not a mechanical engineer, although I shared many undergraduate courses with some of them. I am a nuclear engineer with a strong background on mathematics and computer programming. I went to college between 2002 and 2008. Probably a lot of things have changed since then---at least that is what these millenials guys and girls seem to be boasting about---but chances are we all studied solid mechanics and heat transfer with a teacher using a piece of chalk on a blackboard and students writing down notes with pencils on paper sheets. And there is really not much that one can do with pencil and paper regarding mechanical analysis. Any actual case worth the time of an engineer need to be more complex than an ideal canonical case with analytical solution.

We will be swinging back and forth between a case study about fatigue analysis in piping systems of a nuclear power plant and more generic and even romantic topics related to finite elements and computational mechanics. These latter regressions will not remain just as abstract theoretical ideas. Not only will they be directly applicable to the development of the main case, but they will also apply to a great deal of other engineering problems tackled with the finite element method.

@@ -32,24 +32,24 @@ In the years following Enrico Fermi’s demonstration that a self-sustainable fi

After further years passed by, engineers (probably the same people that forked section\ III) noticed that fatigue in nuclear power plants was not exactly the same as in other piping systems. There were some environmental factors directly associated to the power plant that was not taken into account by the regular ASME code. Again, instead of writing a new code from scratch, people decided to add correction factors to the previously amended body of knowledge. This is how knowledge evolves, and it is this kind of complexities that engineers are faced with during their professional lives. We have to face it, it would be a very hard work to re-write everything from scratch every time something changes.

**figure of a CAD pipe system RO-02 12D-24 33410**
![A real-life piping system.](cad1.png)

## Nuclear reactors

In each of the countries that have at least one nuclear power plant there exists a national regulatory body who is responsible for allowing the owner to operate the reactor. These operating licenses are time-limited, with a range that can vary from 25 to 60 years, depending on the design and technology of the reactor. Once expired, the owner might be entitled to an extension, which the regulatory authority can accept provided it can be shown that a certain (and very detailed) set of safety criteria are met. One particular example of requirements is that of fatigue in pipes, especially those that belong to systems that are directly related to the reactor safety.

## Pressurised pipes
## Pressurised pipes

How come that pipes are subject to fatigue? Well, on the one hand and without getting into many technical details, the most common nuclear reactor design uses liquid water as coolant and moderator. On the other hand, nuclear power plants cannot by-pass the thermodynamics of the Carnot cycle, and in order to maximise the efficiency of the conversion between the energy stored in the uranium nuclei into electricity they need to reach temperatures as high as possible. So, if we want to have liquid water in the core as hot as possible, we need to increase the pressure. The limiting temperature and pressure are given by the [critical point of water](https://en.wikipedia.org/wiki/Critical_point_(thermodynamics)), which is around 374ºC and 22\ MPa. It is therefore expected to have temperature and pressures near those values in many systems of the plant, especially in the primary circuit those that directly interact with it, such as pressure and inventory control system, decay power removal system, feedwater supply system, emergency core-cooling system, etc.
How come that pipes are subject to fatigue? Well, on the one hand and without getting into many technical details, the most common nuclear reactor design uses liquid water as coolant and moderator. On the other hand, nuclear power plants cannot by-pass the thermodynamics of the Carnot cycle, and in order to maximise the efficiency of the conversion between the energy stored in the uranium nuclei into electricity they need to reach temperatures as high as possible. So, if we want to have liquid water in the core as hot as possible, we need to increase the pressure. The limiting temperature and pressure are given by the [critical point of water](https://en.wikipedia.org/wiki/Critical_point_(thermodynamics)), which is around 374ºC and 22\ MPa. It is therefore expected to have temperature and pressures near those values in many systems of the plant, especially in the primary circuit those that directly interact with it, such as pressure and inventory control system, decay power removal system, feedwater supply system, emergency core-cooling system, etc.

Nuclear power plants are not always working at 100% power. They need to be maintained and refuelled, they may undergo operational transients, they might operates at a lower power due to load following conditions, etc. These transient cases involved changes both in temperatures and in pressures that the pipes are subject to, which in turn give rise to changes in the stress tensor of the pipes. As the transients are postulated to occur conservatively cyclically during a number of times during the life-time of the plant (plus its extension period), mechanical fatigue in these piping systems arise especially at the interfaces between materials with different thermal expansion coefficients.
Nuclear power plants are not always working at 100% power. They need to be maintained and refuelled, they may undergo operational transients, they might operates at a lower power due to load following conditions, etc. These transient cases involved changes both in temperatures and in pressures that the pipes are subject to, which in turn give rise to changes in the stresses within the pipes. As the transients are postulated to occur conservatively cyclically during a number of times during the life-time of the plant (plus its extension period), mechanical fatigue in these piping systems arise especially at the interfaces between materials with different thermal expansion coefficients.

## Fatigue {#sec:fatigue}

Mechanical systems can fail due to a wide variety of reasons. The effect known as fatigue can create, migrate and grow microscopic cracks at the atomic level, called dislocations. Once these cracks reach a critical size, then the material fails catastrophically even under stresses lower than tensile strength limits. There are not complete mechanistic models from first principles which can be used in general situations, and those that exist are very complex and hard to use. Instead, using an experimental approach very much like the Hooke Law experiment, the stress amplitude of a periodic cycle can be related to the number of cycles where failure by fatigue is expected to occur. For each material, this dependence can be obtained using normalised tests and a family of “fatigue curves” for different temperatures can be obtained.
Mechanical systems can fail due to a wide variety of reasons. The effect known as fatigue can create, migrate and grow microscopic cracks at the atomic level, called dislocations. Once these cracks reach a critical size, then the material fails catastrophically even under stresses lower than tensile strength limits. There are not complete mechanistic models from first principles which can be used in general situations, and those that exist are very complex and hard to use. Instead, using an experimental approach very much like the Hooke Law experiment, the stress amplitude of a periodic cycle can be related to the number of cycles where failure by fatigue is expected to occur. For each material, this dependence can be computed using normalised tests and a family of “fatigue curves” (also called $S$-$N$ curves) for different temperatures can be obtained.


**fatigue curve**
![A fatigue or $S$-$N$ curve for two steels.](SN.svg)

It should be stressed that the fatigue curves are obtained in a particular load case, namely purely-periodic one-dimensional, which is not directly generalised to other three-dimensional cases. The application of the curve data implies a set of simplifications and assumptions that are translated into different possible “rules” for composing real-life cycles. There also exist two safety factors which increase the stress amplitude and reduce the number of cycles respectively. All these intermediate steps render the analysis of fatigue into a conservative computation scheme. Therefore, when a fatigue analysis performed using the fatigue curve method arrives at the conclusion that “fatigue is expected to occur after ten thousand cycles” what it actually means is “we are sure fatigue will not occur before ten thousand cycles, yet it may not occur before one hundred thousand or even more.”

@@ -64,7 +64,7 @@ So, let us start our journey. Our starting place: undergraduate solid mechanics

We have to accept that there is certain intellectual beauty when complex stuff can be expressed in simple term. Yet, from now on, everything can be complicated at will. We can take the mathematical path like D’Alembert and his virtual displacements ideas (in his mechanical treatise, D’Alembert brags that he does not need to use a single figure throughout the book). Or we can go graphical following Cullman. Or whatever other logic reasoning to end up with a set of actual equations which we need to solve in order to obtain engineering results.

## The stress tensor
## The stress tensor {#sec: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.

@@ -94,20 +94,68 @@ It looks (and works) like a regular $3 \times 3$ matrix. Some brief comments abo

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.
* 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

Let us proceed to a our second step, and consider an infinite pipe subject to uniform internal pressure. Actually, we are going to solve the mechanical problem on an infinite hollow cylinder, which looks like pipe. This case is usually tackled in college courses, and chances are you already solved it. Actually, the first (and simpler) problem is the “thin cylinder problem.” Then, the “thick cylinder problem” is introduced, which is slightly more complex. Nevertheless, it has an analytical solution.
Let us proceed to a our second step, and consider an infinite pipe subject to uniform internal pressure. Actually, we are going to solve the mechanical problem on an infinite hollow cylinder, which looks like pipe. This case is usually tackled in college courses, and chances are you already solved it. Actually, the first (and simpler) problem is the “thin cylinder problem.” Then, the “thick cylinder problem” is introduced, which is slightly more complex. Nevertheless, it has an analytical solution which is derived in [@pipe-linearized]. For the present case, Let us consider an infinite pipe (i.e. a hollow cylinder) of internal radius $a$ and external radius $b$ with uniform mechanical properties---Young modulus $E$ and Poisson’s ratio $\nu$---subject to an internal uniform pressure $p$.

dnl google thin walled pressure vessel strain
### Displacements

**equilibrium equation**
Remember that when any solid body is subject to external forces, it has to reach in such a way to satisfy the equilibrium conditions. The way solids do this is by deforming a little bit in such a way that the whole body acts as a compressed (or elongated) spring balancing the load. So it is worth to ask how a pressurized pipe deforms to counteract the internal pressure\ $p$.

**stresses solution**
* There are no longitudinal displacements\ $u_l$ because the pipe is infinite in the axial direction.
* There are no azimuthal displacements\ $u_\theta$ because the pipe is fully symmetric around the axis.
* There are only radial displacements\ $u_r$ and they depend only on the radial coordinate\ $r$ and not on the axial position\ $z$ or on the azimuthal angle\ $\theta$. This displacements are

$$
u_r(r) = p \cdot \frac{1+\nu}{E} \cdot \frac{a^2}{b^2-a^2} \cdot \left[ 1-2\nu + \frac{b^2}{r^2} \right]\cdot r
$$

What does this mean? Well, that overall the whole pipe expands a little bit radially with the inner face being displaced more than the external surface (use your imagination!). How much?

1. Linearly with the pressure, i.e. twice the pressure, twice the displacement, and
2. Inversely proportional to the Young Modulus\ $E$ divided by $1+\nu$, i.e. the more resistant the material, the less radial displacements.

That is how an infinite pipe withstands internal pressure.

### Stresses

As the solid is deformed, that is to say that different parts are relatively displaced one from another, strains and stresses appear. When seen from a cylindrical coordinate system, the stress tensor (recall [@sec:tensor]) has these features.

* There are no shear stresses as there is no bending due to the fact that the pipe is infinite (so it cannot bend in the axial direction) and azimuthally symmetric (there is no particular direction so circles must remain circles).
* The normal stresses depend only on the radial coordinate\ $r$ and are
- the radial stress\ $\sigma_r$,

$$ \sigma_r(r) = \frac{p \cdot a^2}{b^2-a^2} \cdot \left( 1 - \frac{b^2}{r^2}\right) $$ {#eq:sigmar}

- the azimuthal (or hoop) stress\ $\sigma_\theta$, and

$$ \sigma_\theta(r) =\frac{p \cdot a^2}{b^2-a^2} \cdot \left( 1 + \frac{b^2}{r^2}\right) $$ {#eq:sigmatheta}

- the longitudinal (or axial) stress\ $\sigma_l$.

$$ \sigma_l(r) = 2\nu \cdot \frac{p \cdot a^2}{b^2-a^2} $${#eq:sigmal}

We can note that

1. The stresses do not depend on the mechanical properties\ $E$ and\ $\nu$ of the material (the displacements do).
2. All the stresses are linear with the pressure\ $p$, i.e. twice the pressure, twice the stress.
3. The axial stress is uniform and does not depend on the radial coordinate\ $r$.
4.
That is all what we can say about an infinite pipe with uniform material properties subject to an uniform internal pressure\ $p$. If
* the pipe was not infinite (say any real pipe that has to start and end somewhere), or
* the cross-section of the pipe is not constant along the axis (say there is a reduction), or
* there was more than one pipe (say there is a tee), or
* the material properties are not uniform (say the pipe does not have an uniform temperature but a distribution), or
* the pressure was not uniform (say because there is liquid inside and its weight cannot be neglected),
\noindent then we would no longer be able to fully solve the problem with paper and pencil and draw all the conclusions above. However, at least we have a start because we know that if the pipe is finite but long enough or the temperature is not uniform but almost, we still can use the analytical equations as approximations. But what happens is the pipe is short, there are branches and temperature changes like during a transient in a nuclear reactor? Well, that is why we have finite elements.


# Finite elements, or solving an actual engineering problem

@@ -122,18 +170,18 @@ There are literally dozens of ways to numerically solve the equilibrium equation
1. Finite differences
2. Finite volumes
3. Finite elements
Each of these methods (also called schemes) have of course their own features, pros and cons. They all exploit the fact that the equations are easy to solve in simple geometries (say a cube). Then the actual geometry is divided into a yuxtaposition of these cubes, the equations are solved in each one and then a global solution is obtained by sewing the little simple solutions one to another. The process of dividing the original domain into simple geometries is called _discretization_, and the resulting collection of these simple geometries is called a mesh or grid. They are composed of volumes, called cells (or elements) and vertices called nodes. Now, grids can be either

a. structured, or
b. unstructured
[Figure@fig:grids] illustrate how the same domain can be discretized using these two kind of grids. In the first case, we could identify any single cell by using just two indexes. We could even tell which nodes define each cell just from these indexes. In the second case, we need an explicit list first to know how many cells there are. Even more, there is no way to link the nodes with the cells (back and forth) other than having a list of nodes and cells. Again, there are pros and cons for each of the grid types such as simplicity, flexibility, etc. In general, unstructured grids and better represent a certain geometry with the same number of cells. Structured grids suffer the so-called “staircase effect” that makes the unusable for discretizing mechanical parts.

The first of the three methods is based on approximating derivative (i.e. differentials) by incremental quotients (i.e. differences). The second one heavily relies on geometrical ideas rather than on pure mathematical grounds. Finally, our beloved finite elements are the most “mathematical” ones. Actually, a complete derivation of the finite element method can be written in a textbook without requiring a single figure, just like D’Alembert did more than two centuries ago. In any case, it is important to note that finite differences and elements compute results at the _nodes_ of a mesh, whilst finite volumes compute results at the _cells_ of a mesh. Finally, any method may be used in structured grids but only finite elements and volumes are especially suited for working with unstructured grids.

There are technical reasons that justify why the finite element method is the king of mechanical analysis. But that does not mean that other methods may be employed. For instance, fluid mechanics are better solved using finite volumes. And further other combinations may be found in the literature.
The first of the three methods is based on approximating derivative (i.e. differentials) by incremental quotients (i.e. differences). The second one heavily relies on geometrical ideas rather than on pure mathematical grounds. Finally, our beloved finite elements are the most “mathematical” ones. Actually, a complete derivation of the finite element method can be written in a textbook without requiring a single figure, just like D’Alembert did more than two centuries ago. In any case, it is important to note that finite differences and elements compute results at the _nodes_ of a mesh, whilst finite volumes compute results at the _cells_ of a mesh. Finally, any method may be used in structured grids but only finite elements and volumes are especially suited for working with unstructured grids.

There are technical reasons that justify why the finite element method is the king of mechanical analysis. But that does not mean that other methods may be employed. For instance, fluid mechanics are better solved using finite volumes. And further other combinations may be found in the literature.

Before proceeding, I would like to make two comments about common nomenclature. The first one is that if we exchanged the words “volumes” and “elements” in all the written books and articles, nobody would note the difference. There is nothing particular in both theories that can justify why finite volumes use volumes and finite elements use elements. Actually volumes and elements are the same geometric constructions. The names were randomly assigned.

@@ -143,7 +191,7 @@ The second one is more philosophical and refers to the word “simulation” whi

This section is not (just) about different kinds of elements like tetrahedra, hexahedra, pyramids and so on. It is about the different kinds of analysis there are. Indeed, there are a whole plethora of particular types of calculations we can perform, all of which can be called “finite element analysis.” For instance, for the mechanical problem, we can have different kinds of

* temporal dependence
* temporal dependence
- steady-state
- quasi-static
- transient
@@ -168,15 +216,15 @@ And then there exist different pre-processors, meshers, solvers, pre-conditioner
i. conservative
ii. best-estimate
iii. probabilistic
The first one is the easiest because we are allowed to choose parameters and to make engineering decisions that may simplify the computation as long as they give results towards the worse-case scenario. More often than not, an conservative _estimation_ is enough in order to consider a problem solved. Note that this is actually how fatigue results are obtained using fatigue curves, as discussed in\ [@sec:fatigue]. A word of care should be taken when considering what the “worst-case scenario” is. For instance, if we are analysing the temperature distribution in a mechanical part subject to convection boundary conditions, we might take either a very large or a very low convection coefficient as the conservative case. If we needed to design fins to dissipate heat then a low coefficient would be the choice conservative. But if the mechanical properties deteriorated with high temperatures then the conservative way to go would be to set a high convection coefficient. A common practice is to have a fictitious set of parameters, each of them being conservative leading individually to the worst-case scenario even if the overall combination is not physically feasible.

As neat and tempting as conservative computations may be, sometimes the assumptions may be too biased toward the worst-case scenario and there might be no way of justifying certain designs with conservative computations. It is then time to sharpen our pencils and perform a best-estimate computation. This time, we should stick to the most-probable values of the parameters and even use more complex models that can better represent the physical phenomena that are going on in our problem. Sometimes best-estimate computations are just slightly more complex than conservative models. But more often than not, best-estimates get far more complicated. And these complications come not just in the finite-element model of the elastic problem but in the dependence of properties with space, time and/or temperature, in non-trivial relationships between macro and microscopic parameters, in more complicated algorithms for post-processing data, etc.
As neat and tempting as conservative computations may be, sometimes the assumptions may be too biased toward the worst-case scenario and there might be no way of justifying certain designs with conservative computations. It is then time to sharpen our pencils and perform a best-estimate computation. This time, we should stick to the most-probable values of the parameters and even use more complex models that can better represent the physical phenomena that are going on in our problem. Sometimes best-estimate computations are just slightly more complex than conservative models. But more often than not, best-estimates get far more complicated. And these complications come not just in the finite-element model of the elastic problem but in the dependence of properties with space, time and/or temperature, in non-trivial relationships between macro and microscopic parameters, in more complicated algorithms for post-processing data, etc.

**Example?**

Finally, when then uncertainties associated to the parameters, methods and models used in a best-estimate calculation render the results too inaccurate for a certain regulatory body to approve a design, it might be needed to do a full set of parametric runs taking into account the probabilistic distribution of each of the input parameters. This kind of computation involve
1. a thorough analysis of the probability densities of the parameters (and even the methods) of a problem,
2. performing a large number of runs for different combination of parameters, and
3. combining all the results into to obtain a best-estimate value plus uncertainty.
@@ -196,28 +244,28 @@ Here is an [original example](https://www.toyota-global.com/company/toyota_tradi

1. Why did the robot stop?
The circuit has overloaded, causing a fuse to blow.
2. Why is the circuit overloaded?
There was insufficient lubrication on the bearings, so they locked up.
3. Why was there insufficient lubrication on the bearings?
The oil pump on the robot is not circulating sufficient oil.
4. Why is the pump not circulating sufficient oil?
The pump intake is clogged with metal shavings.
5. Why is the intake clogged with metal shavings?
Because there is no filter on the pump.

You get the point. We usually assume we have to do what we usually do (i.e. perform finite element analysis). But do we? Do we add a filter or just replace the fuse?

Getting back to the case study: do we need to do FEM analysis? Well, it does not look like we can obtain the stresses the transient cases with just pencil and paper.
Getting back to the case study: do we need to do FEM analysis? Well, it does not look like we can obtain the stresses the transient cases with just pencil and paper.


divert(-1)
## Computers, those little magic boxes

When we think about finite elements, we automatically think about computers. Of
When we think about finite elements, we automatically think about computers. Of

https://www.springfieldspringfield.co.uk/view_episode_scripts.php?tv-show=the-simpsons&episode=s05e03


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