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gtheler 6 anni fa
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16 ha cambiato i file con 2275 aggiunte e 371 eliminazioni
Visualizzazione separata Mostra Diff Stats
  1. +12
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      analytical.ppl
  2. +7
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      convergence-1.dat
  3. +7
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      convergence-2.dat
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      error-M-vs-cpu-small.pdf
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      error-M-vs-cpu-small.pdf.pdf
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      error-M-vs-cpu-small.svg
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      error-MB-vs-cpu-small.pdf
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      error-MB-vs-cpu-small.pdf.pdf
  9. +347
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      error-MB-vs-cpu-small.svg
  10. +31
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      error.ppl
  11. +3
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      meta.yaml
  12. +1356
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      nafems4-full.md
  13. +150
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      nafems4.md
  14. +18
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      problem.ppl
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      real-gen.png
  16. +1
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      upload.sh

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analytical.ppl Vedi File

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# analytical results in [ MPa ]
M_tresca = 69.9467
M_vonmises = 61.7785
M_1 = 65.3023
M_2 = 18.1974
M_3 = -4.64439

MB_tresca = 79.9062
MB_vonmises = 70.2562
MB_1 = 70.2821
MB_2 = 18.1974
MB_3 = -9.62415

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convergence-1.dat Vedi File

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# FEM with order 1 elements
1 68.704 60.700 64.380 18.013 -4.324 70.314 62.277 67.777 19.568 -6.111 276 915 0.02 29
2 69.461 61.361 65.010 18.175 -4.451 75.171 66.329 68.817 18.522 -6.354 1544 6431 0.15 51
3 69.416 61.343 65.195 18.277 -4.221 77.030 67.858 69.489 18.542 -7.541 3765 17215 0.49 88
4 69.575 61.462 65.073 18.161 -4.503 78.182 68.806 69.679 18.343 -8.503 7351 34855 1.53 146
5 69.649 61.531 65.143 18.161 -4.507 78.280 68.880 69.548 18.214 -8.732 12892 64457 3.30 242
6 70.033 61.853 65.348 18.194 -4.685 79.435 69.859 70.143 18.264 -9.292 20984 108170 6.39 386

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convergence-2.dat Vedi File

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# FEM with order 2 elements
1 69.763 61.606 65.001 18.075 -4.762 79.996 70.319 70.127 18.079 -9.869 1389 915 0.29 46
2 69.932 61.762 65.257 18.176 -4.675 79.904 70.251 70.227 18.188 -9.677 9128 6431 3.79 282
3 69.959 61.787 65.289 18.187 -4.670 79.891 70.241 70.254 18.188 -9.637 23816 17215 13.42 671
4 69.904 61.742 65.273 18.192 -4.631 79.861 70.217 70.248 18.195 -9.613 48089 34855 26.24 1252
5 69.923 61.758 65.286 18.195 -4.637 79.883 70.236 70.264 18.196 -9.619 87900 64457 58.79 2275
6 69.922 61.758 65.288 18.196 -4.635 79.933 70.278 70.293 18.196 -9.640 146766 108170 121.44 3791

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error-M-vs-cpu-small.pdf Vedi File


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error-M-vs-cpu-small.pdf.pdf Vedi File


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error-M-vs-cpu-small.svg Vedi File

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+ 31
- 0
error.ppl Vedi File

@@ -0,0 +1,31 @@
load "problem.ppl";
load "analytical.ppl"

set preamble "\usepackage{amsmath}"

set width 7*unit(cm)

set axis x arrow nomirrored
set axis y arrow nomirrored


set grid
set terminal pdf

set output "error-M-vs-cpu-small.pdf"
set key above

set yrange [1e-3:1e2]
set logscale y
set logscale x

set xlabel "CPU time [sec]"
set ylabel "Error in M [MPa]"
plot "convergence-1.dat" u 14:(abs($3-M_vonmises)) w lp lw 2 lt 3 pt 16 color orangered ti "First-order",\
"convergence-2.dat" u 14:(abs($3-M_vonmises)) w lp lw 3 lt 4 pt 17 color cadetblue ti "Second-order"

set output "error-MB-vs-cpu-small.pdf"
set xlabel "CPU time [sec]"
set ylabel "Error in MB [MPa]"
plot "convergence-1.dat" u 14:(abs($8-MB_vonmises)) w lp lw 2 lt 3 pt 16 color salmon ti "First-order",\
"convergence-2.dat" u 14:(abs($8-MB_vonmises)) w lp lw 3 lt 4 pt 17 color navyblue ti "Second-order"

+ 3
- 5
meta.yaml Vedi File

@@ -1,8 +1,6 @@
---
title: Fatigue Assesment in Nuclear Piping
subtitle: A journey from college theory to an actual engineering problem
author: Jeremy Theler
authoremail: jeremy@seamplex.com
title: The NAFEMS Benchmark Challenge Volume\ 1
author: Angus Ramsay
fontsize: 10pt
lang: en-GB
bibliography: references.bib
@@ -15,7 +13,7 @@ mathspec: true
formats: html tex pdf epub docx odt
listings: true
template: simple
documentclass: article
documentclass: book
classoption: twoside
geometry: "a5paper, left=15mm, right=15mm, bottom=25mm, top=20mm, foot=15mm, head=15mm"
...

+ 1356
- 0
nafems4-full.md
File diff soppresso perché troppo grande
Vedi File


+ 150
- 365
nafems4.md Vedi File

@@ -1,4 +1,15 @@
# Background and introduction
# Introduction

# CS1: Design of a Bandsaw Pulley

# CS2: Torsion of Curved Beams

# CS3: Design of Steel Silos


# CS4: Fatigue Assessment in Nuclear Piping

## From college theory to an actual engineering problem

First of all, please take this text as a written chat between you an me, i.e. an average engineer that has already taken the journey from college to performing actual engineering works 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.

@@ -30,7 +41,7 @@ Finite elements are like magic to me. I mean, I can follow the whole derivation
Again, take all this information as coming from a fellow that has already taken such a journey from college’s pencil and paper to real engineering cases involving complex numerical calculations. And developing, in the meantime, both an actual working finite-element [back-end](https://www.seamplex.com/fino) and [front-end](https://www.caeplex.com) from scratch.^[The dean of the engineering school I attended used to say “It is not the same to read than to write manuals, and we should aim at writing them.”]


## Tips and tricks
### Tips and tricks

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.

@@ -47,7 +58,7 @@ So here comes another experience tip: do not fear equations. Even more, keep exe
One final comment: throughout the text I will be referring to “your favourite FEM program.” I bet you do have one. Mine is [CAEplex](https://caeplex.com) (it works on top of [Fino](https://www.seamplex.com/fino)). We will be using it to perform some tests and play a little bit. And we will also use it to think about what it means to use a FEM program to generate results that will eventually end up in a written project with your signature. Keep that in mind.


# Case study: reactors, pipes and fatigue {#sec:case}
## Case study: reactors, pipes and fatigue {#sec:case}

Piping systems in sensitive industries like nuclear or oil & gas should be designed and analysed following the recommendations of an appropriate set of codes and norms, such as the [ASME\ Boiler and Pressure Vessel Code](https://en.wikipedia.org/wiki/ASME_Boiler_and_Pressure_Vessel_Code).
This code of practice was born during the late\ XIX century, before finite-element methods for solving partial differential equations were even developed. And much longer before they were available for the general engineering community. Therefore, much of the code assumes design and verification is not necessarily performed numerically but with paper and pencil (yes, like in college). However, it still provides genuine guidance in order to ensure pressurised systems behave safely and properly without needing to resort to computational tools. Combining finite-element analysis with the ASME code gives the cognisant engineer a unique combination of tools to tackle the problem of designing and/or verifying pressurised piping systems.
@@ -58,13 +69,13 @@ After further years passed by, engineers (probably the same people that forked s

Actually, this article does not focus on a single case study but on some general ideas regarding analysis of fatigue in piping systems in nuclear power plants. There is no single case study but a compendium of ideas obtained by studying many different systems which are directly related to the safety of a real nuclear reactor.

![Three-dimensional CAD model for the piping system of\ [@fig:isometric]](real-piping.png){#fig:real-life}
![Three-dimensional CAD model for the piping system of\ [@fig:isometric]](real-piping.png){#fig:real-life width=75%}

## Nuclear reactors
### 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](https://en.wikipedia.org/wiki/Carnot_cycle), and in order to maximise the efficiency of the conversion of the energy stored in the uranium nuclei into electricity we 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 temperatures and pressures near those values in many systems of the plant, especially in the primary circuit and 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.
@@ -73,7 +84,7 @@ Nuclear power plants are not always working at 100% of their maximum power capac

An important part of the analysis that almost always applies to nuclear power plants but usually also to other installations is the consideration of a possible seismic event. Given a postulated design earthquake, both the civil structures and the piping system itself need to be able to withstand such a load, even if it occurs at the moment of highest mechanical demand during one of the operational transients.

## Fatigue {#sec:fatigue}
### 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](https://en.wikipedia.org/wiki/Dislocations). Once these cracks reach a critical size, then the material fails catastrophically even under stresses much lower than the 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. There are two main ways to approach practical fatigue assessment problems using experimental data very much like the [Hooke’s Law](https://en.wikipedia.org/wiki/Hooke's_law) experiment:

@@ -85,11 +96,11 @@ The first one is suitable for cases where the loads are nowhere near the yield s
For the case study, as the loads come principally from operational loads, the ASME\ stress-life approach should be used. 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” like the one depicted in\ [@fig:SN] (also called $S$-$N$ curve) for different temperatures can be obtained.


![A fatigue or $S$-$N$ curve for two steels.](SN.svg){#fig:SN width=95%}
![A fatigue or $S$-$N$ curve for two steels.](SN.svg){#fig:SN width=75%}

It should be noted that the fatigue curves are obtained in a particular load case, namely purely-periodic and one-dimensional, which cannot be directly generalised to other three-dimensional cases. Also, any real-life case will be subject to a mixture of complex cycles given by a stress time history and not to pure periodic conditions. 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 ([@sec:kinds]). Therefore, when a fatigue assessment 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.”

# Solid mechanics, or what we are taught at college
## Solid mechanics, or what we are taught at college

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](https://en.wikipedia.org/wiki/Newton's_laws_of_motion). For our purposes, we can recall them here like this:

@@ -99,7 +110,7 @@ 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 such simple terms. Yet, from now on, everything can be complicated at will. We can take the mathematical path like [D’Alembert](https://en.wikipedia.org/wiki/Jean_le_Rond_d'Alembert) and his virtual displacements ideas (in his mechanical treatise, he brags that he does not need to use a single figure throughout the book). Or we can go graphical following [Culmann](https://en.wikipedia.org/wiki/Carl_Culmann). 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 {#sec: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 mathematics. Trust me, it will pay back later on.

@@ -119,12 +130,7 @@ It looks (and works) like a regular $3 \times 3$ matrix. Some brief comments abo

* 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 twist the material.
* Due to rotational equilibrium requirements the conjugate shear stresses should be equal:
- $\tau_{xy} = \tau_{yx}$,
- $\tau_{yz} = \tau_{zy}$, and
- $\tau_{zx} = \tau_{xz}$.

Therefore, the stress tensor is symmetric i.e. there are only six independent elements.
* Due to rotational equilibrium requirements the conjugate shear stresses should be equal: $\tau_{xy} = \tau_{yx}$, $\tau_{yz} = \tau_{zy}$, and $\tau_{zx} = \tau_{xz}$. Therefore, the stress tensor is symmetric i.e. there are only six independent elements.
* The elements of the tensor depend 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](https://en.wikipedia.org/wiki/Principal_stresses), which happen to be the three [eigenvalues](https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors) of the stress tensor. The basis of the coordinate system that renders the tensor diagonal are the eigenvectors.

@@ -134,11 +140,11 @@ What does this all have to do with mechanical engineering? Well, once we know wh
> If you can compute the stress tensor at each point of your geometry, then... Congratulations! You have solved the solid mechanics problem.


## An infinitely-long pressurised pipe {#sec:infinite-pipe}
### An infinitely-long pressurised pipe {#sec:infinite-pipe}

Let us proceed to our second step, and consider the infinite pipe subject to uniform internal pressure already introduced in\ [@fig:infinite-pipe]. 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. In fact, the first (and simpler) problem is the “thin cylinder problem.” Then, the “thick cylinder problem” is introduced (the one we solve below), which is slightly more complex. Nevertheless, it has an analytical solution which is derived [here](https://www.seamplex.com/fino/doc/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 $p$.

### Displacements
#### Displacements

Remember that when any solid body is subject to external forces, it has to react 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 pressurised pipe deforms to counteract the internal pressure\ $p$.

@@ -157,23 +163,15 @@ What does this mean? Well, that overall the whole pipe expands a little bit radi

That is how an infinite pipe withstands internal pressure.

### Stresses
#### 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}
- the radial stress $\sigma_r(r) = \frac{p \cdot a^2}{b^2-a^2} \cdot \left( 1 - \frac{b^2}{r^2}\right)$
- the azimuthal (or hoop) stress $\sigma_\theta(r) =\frac{p \cdot a^2}{b^2-a^2} \cdot \left( 1 + \frac{b^2}{r^2}\right)$, and
- the longitudinal (or axial) stress $\sigma_l(r) = 2\nu \cdot \frac{p \cdot a^2}{b^2-a^2}$

We can note that

@@ -194,11 +192,11 @@ That is all what we can say about an infinite pipe with uniform material propert



# Finite elements, or solving actual problems
## Finite elements, or solving actual problems

Besides infinite pipes (both thin and thick), spheres and a couple of other geometries, there are no other cases for which we can obtain analytical expressions for the elements of the stress tensor. To get results for a solid with real engineering interest, we need to use numerical methods to solve the equilibrium equations. It is not that the equations are hard _per se_. It is that the mechanical parts we engineers like to design (which are of course more complex than cylinders and spheres) are so intricate that render simple equations into monsters which are unsolvable with pencil and paper. Hence, finite elements enter into the scene.

## The name of the game {#sec:formulations}
### The name of the game {#sec:formulations}

But before turning our attention directly into finite elements (and leaving college, at least undergraduate) it is worth some time to think about other alternatives. Are we sure we are tackling your problems in the best possible way? I mean, not just engineering problems. Do we take a break, step back for a while and see the whole picture looking at all the alternatives so we can choose the best cost-effective one?

@@ -208,6 +206,7 @@ There are literally dozens of ways to numerically solve the equilibrium equation
2. [Finite volumes](https://en.wikipedia.org/wiki/Finite_volume_method)
3. [Finite elements](https://en.wikipedia.org/wiki/Finite_element_method)

divert(-1)
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 juxtaposition 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 _discretisation_, 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
@@ -228,41 +227,24 @@ Back to the three numerical methods, we must say that finite differences is base

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 generally better solved using finite volumes. And further other combinations may be found in the literature.

divert(0)

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 notice the difference. There is nothing particular in both theories that can justify why FVM use “volumes” and FEM use “elements”. Actually volumes and elements are the same geometric constructions. As far as I know, the names were randomly assigned.

The second one is more philosophical and refers to the word “simulation” which is often used to refer to solving a problem using a numerical scheme such as the finite element method. [I am against at using this word for this endeavour](https://www.seamplex.com/blog/say-modeling-not-simulation.html). The term simulation has a connotation of both “pretending” and “faking” something, that is definitely not what we are doing when we solve an engineering problem with finite elements. Sure, there are some cases in which we simulate, such as using the [Monte Carlo method](https://en.wikipedia.org/wiki/Monte_Carlo_method) (originally used by Fermi as an attempt to understand how neutrons behave in the core of nuclear reactors). But when solving deterministic mechanical engineering problems I would rather say “modelling” than “simulation.”

## Kinds of finite elements {#sec:kinds}
### Kinds of finite elements {#sec:kinds}

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 is 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
- steady-state
- quasi-static
- transient
* main elements
- 1D beam elements
- 2D shell elements
- 3D bulk elements
* mathematical models
- pure linear
- material non-linear
- geometrical non-linear
* particular studies
- buckling
- modal
* element features
- isoparametric elements
- serendipity elements
- sub-integrated elements
- incomplete elements

And then there exist different pre-processors, meshers, solvers, pre-conditioners, post-processing steps, etc. A similar list can be made for the [heat conduction problem](https://en.wikipedia.org/wiki/Thermal_conduction), [electromagnetism](https://en.wikipedia.org/wiki/Electromagnetism), the [Schröedinger equation](https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation), [neutron transport](https://en.wikipedia.org/wiki/Neutron_transport), etc. But there is also another level of “kind of problem,” which is related to how much accuracy and precision we are to willing sacrifice in order to have a (probably very much) simpler problem to solve. Again, there are different combinations here but a certain problem can be solved using any of the following three approaches, listed in increasing amount of difficulty and complexity:

i. conservative
ii. best-estimate
iii. probabilistic
* temporal dependence: steady-state, quasi-static, transient, ...
* main elements: 1D beam elements, 2D shell elements, 3D bulk elements, ...
* mathematical models: pure linear, material non-linear, geometrical non-linear, particular studies, buckling, modal, ...
* element features: isoparametric elements, serendipity elements, sub-integrated elements, incomplete elements, ...

And then there exist different pre-processors, meshers, solvers, pre-conditioners, post-processing steps, etc. A similar list can be made for the [heat conduction problem](https://en.wikipedia.org/wiki/Thermal_conduction), [electromagnetism](https://en.wikipedia.org/wiki/Electromagnetism), the [Schröedinger equation](https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation), [neutron transport](https://en.wikipedia.org/wiki/Neutron_transport), etc. But there is also another level of “kind of problem,” which is related to how much accuracy and precision we are to willing sacrifice in order to have a (probably very much) simpler problem to solve. Again, there are different combinations here but a certain problem can be solved using any of the following three approaches, listed in increasing amount of difficulty and complexity: conservative, best-estimate or probabilistic.

divert(-1)
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 worst-case scenario. More often than not, a conservative _estimation_ is enough in order to consider a problem as 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 conservative choice. 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 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 bad direction 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.
@@ -274,9 +256,12 @@ Finally, when then uncertainties associated to the parameters, methods and model
3. combining all the results into to obtain a best-estimate value plus uncertainty.

This kind of computation is usually required by the nuclear regulatory authorities when power plant designers need to address the safety of the reactors. What is the heat capacity of uranium above 1000ºC? What is the heat transfer coefficient when approaching the [critical heat flux](https://en.wikipedia.org/wiki/Critical_heat_flux) before the [Leidenfrost effect](https://en.wikipedia.org/wiki/Leidenfrost_effect) occurs? A certain statistical analysis has to be done prior to actually parametrically sweeping (see\ [@sec:parametric]) the input parameters so as to obtain a distribution of possible outcomes.
divert(0)

We might get into a an infinite taxonomic loop if we continue down this path. So let us move one step closer to our case study in this journey from college theory to an actual engineering problem.

## Five whys {#sec:five}
divert(-1)
### Five whys {#sec:five}

So we know we need a numerical scheme to solve our mechanical problem because anything slightly more complex than an infinite pipe does not have analytical solution. We need an unstructured grid because we would not use Legos to discretise cylindrical pipes. We selected the finite elements method over the finite volumes method, because FEM is the king. Can we pause again and ask ourselves why is it that we want to do finite-element analysis?

@@ -313,47 +298,9 @@ Getting back to the case study: do we need to do FEM analysis? Well, it does not
3. to analyse the output data and write engineering reports.

In the first years of the [history of computers](https://en.wikipedia.org/wiki/History_of_computing_hardware), when programs were written in decks and output results were printed in continuous paper sheets, it made sense for computer programs to calculate and write as much data as possible even if it was not needed. One would never know if it would not be needed in the future, and CPU time was so expensive that re-running engineering computations because a particular result was not included in the output was forbidden. But that is not remotely true in the XXI century anymore. Computing time is far cheaper than engineering time (result known as the [UNIX Rule of Economy](http://www.catb.org/~esr/writings/taoup/html/ch01s06.html)) that it should be neglected with respect to the time spent by a cognisant engineer searching and sorting thousands of hard-to-read floating-point numbers.

divert(-1)

-----------------------------------------------------------

## Computers, those little magic boxes

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

ENIAC

### A brief review of history

FEM, Computers

graphics cards

### Hardware

### Software

FOSS

Avoid black boxes

Reflections on trusting trust

UNIX, scriptability, make programs to make programs (here a program is a calculation)

front and back

avoid monolithic

-----------------------------------------------------------

divert(0)


# Piping in nuclear rectors {#sec:piping-nuclear}
## Piping in nuclear rectors {#sec:piping-nuclear}

So we need to address the issue of fatigue in nuclear reactor pipes that

@@ -365,7 +312,8 @@ So we need to address the issue of fatigue in nuclear reactor pipes that
b. heat transients, and
c. seismic loads.

As I wanted to illustrate in [@sec:five], it is very important to decide what kind of problem (actually problems) we should be dealing with. As a nuclear engineer, I learned (theoretically in college but practically after college) that there are some models that let you see some effects and some that let you see other effects (please [say “modelling” not “simulation.”](https://www.seamplex.com/blog/say-modeling-not-simulation.html)). And even if, in principle, it is true that more complex models should let you see more stuff, they definitely might show you nothing at all if the model is so big and complex that it does not fit into a computer (say because it needs hundreds of gigabytes of RAM to run) or because it takes more time to compute than you may have before the final report is expected.
dnl As I wanted to illustrate in [@sec:five], it is very important to decide what kind of problem (actually problems) we should be dealing with.
As a nuclear engineer, I learned (theoretically in college but practically after college) that there are some models that let you see some effects and some that let you see other effects (please [say “modelling” not “simulation.”](https://www.seamplex.com/blog/say-modeling-not-simulation.html)). And even if, in principle, it is true that more complex models should let you see more stuff, they definitely might show you nothing at all if the model is so big and complex that it does not fit into a computer (say because it needs hundreds of gigabytes of RAM to run) or because it takes more time to compute than you may have before the final report is expected.

First of all, we should note that we need to solve

@@ -379,33 +327,23 @@ So for each time\ $t$ of the operational transient, the pipes are subject to
b. a non-uniform internal temperature $T_i(t)$ that gives rise to a non-trivial time-dependent temperature distribution\ $T(\mathbf{x},t)$ in the bulk of the pipes, and
c. internal distributed forces\ $\mathbf{f}=\rho \cdot \mathbf{a}$ at those times where the design earthquake is assumed to occur.

## Thermal transient {#sec:thermal}
### Thermal transient {#sec:thermal}

Let us invoke our imagination once again. Assume in one part of the transients the temperature of the water inside the pipes falls from say 300ºC down to 100ºC in a couple of minutes, stays at 100ºC for another couple of minutes and then gets back to 300ºC. The temperature within the bulk of the pipes changes as times evolves. The internal wall of the pipes follow the transient temperature (it might be exactly equal or close to it through the [Newton’s law of cooling](https://en.wikipedia.org/wiki/Newton%27s_law_of_cooling)). If the pipe was in a state of uniform temperature, the ramp in the internal wall will start cooling the bulk of the pipe creating a transient thermal gradient. Due to thermal inertia effects, the temperature can have a non-trivial dependence when the ramps start or end (think about it!). So we need to compute a real transient heat transfer problem with convective boundary conditions because any other usual tricks like computing a sequence of steady-state computations for different times would not be able to recover these non-trivial distributions.

Remember the main issue of the fatigue analysis in these systems is to analyse what happens around the location of changes of piping classes where different materials (i.e. different expansion coefficients) are present, potentially causing high stresses due to differential thermal expansion (or contraction) under transient conditions. Therefore, even though we are dealing with pipes we cannot use beam or circular shell elements, because we need to take into account the three-dimensional effects of the temperature distribution along the pipe thickness. And even if we could, there are some tees that connect pipes with different nominal diameters that have a non-trivial geometry, such as the weldolet-type junction shown in\ [@fig:weldolet-cad;@fig:weldolet-mesh]. In this case, there are a number of SCLs (Stress Classification Lines) that go through the pipe’s thickness at both sides of the material interface as illustrated in\ [@fig:weldolet-scls]. It is in these locations that fatigue is to be evaluated.

dnl 33410 07-3-4D-29

::::: {#fig:weldolet-cad}
![Overall view](weldolet-cad1.png){#fig:weldolet-cad1 width=80%}
::::: {#fig:weldolet}
![Overall view](weldolet-cad1.png){#fig:weldolet-cad width=65%}

![Detail of the weldolet-type junction](weldolet-cad2.png){#fig:weldolet-cad2}
![Unstructured tetrahedra-based grid](weldolet-mesh2.png){#fig:weldolet-mesh width=85%}

CAD model of a piping system with a 3/4-inch weldolet-type fork (stainless steel) from a main 12-inch pipe (carbon steel)
:::::


::::: {#fig:weldolet-mesh}
![Overall view](weldolet-mesh1.png){#fig:weldolet-mesh1 width=55%}

![Detail of the grid around the junction, showing mesh refinement around the material interface (purple and green elements).](weldolet-mesh2.png){#fig:weldolet-mesh2}

Three-dimensional unstructured tetrahedra-based grid for the system shown in\ [@fig:weldolet-cad]
:::::


![Location of the six SCLs defined to analyse fatigue around the junction.](weldolet-scls-n.png){#fig:weldolet-scls width=75%}
![Location of six SCLs defined to analyse fatigue around a junction.](weldolet-scls-n.png){#fig:weldolet-scls width=75%}


On the one hand, a reasonable number of nodes (it is the number of nodes that defines the problem size, not the number of elements as discussed in [@sec:elements-nodes]) in order to get a decent grid is around 200k for each system. On the other hand, solving a couple of dozens of transient heat transfer problems (which we cannot avoid due to the large thermal inertia of the pipes) during a few thousands of seconds over a couple hundred of thousands of nodes might take more time and storage space to hold the results than we might expect.
@@ -417,60 +355,35 @@ We can then merge this idea by Asimov with an adapted version of the [Saint-Vena
1. compute the transient thermal problem using a reduced mesh around the SCLs, and
2. assume the part of the full system which is not contained in the reduced mesh is at an uniform (though not constant) temperature equal to the average of the inner and outer temperatures at each side of the reduced mesh.

dnl 33300 02-D-3-4


::::: {#fig:valve}
![Full CAD model.](valve-cad1.png){#fig:valve-cad1 width=75%}

![Full mesh, approximately 220k nodes.](valve-mesh1.png){#fig:valve-mesh1 width=75%}

![Location of the stress classification lines.](valve-scls1-n.png){#fig:valve-scls1 width=60%}

An example case where the SCLs are located around the junction between stainless-steel valves and carbon steel pipes at both sides of the material interface in the vertical plane both at the top and at the bottom of the pipe
:::::



As an example, let us consider the system depicted in\ [@fig:valve] where there is a stainless-carbon steel interface at the discharge of the valves. Instead of solving the transient heat-conduction problem with the internal temperature of the pipes equal to the temperature of the water in the reference transient condition of the power plant and an external condition of natural convection to the ambient temperature in the whole mesh of\ [@fig:valve-mesh1], a reduced model consisting of half of one of the two valves and a small length of the pipes at both the valve inlet and outlet is used. Once the temperature distribution\ $\hat{T}(\mathbf{x},t)$ for each time is obtained in the reduced mesh ([@fig:valve-temp], which has the origin at the centre of the valve), the actual temperature distribution\ $T(\mathbf{x},t)$ is computed by an algebraic generalisation of $\hat{T}(\mathbf{x},t)$ in the full coordinate system (where the origin is shown in\ [@fig:valve-cad1]). As stated above, those locations which are not covered by the reduced model are generalised with a time-dependent uniform temperature which is the average of the inner and outer temperatures at the inlet and outlet of the reduced mesh. The result is illustrated in figure\ [@fig:valve-gen].

::::: {#fig:valve-temp-gen}
![Reduced mesh around the valve refined around the interface where the transient heat conduction problem is solved.](valve-temp.png){#fig:valve-temp width=80%}

![Generalisation of the resulting temperature to the original mesh from figure [@fig:valve-mesh1]](valve-gen.png){#fig:valve-gen width=95%}

Computation of the thermal problem in a reduced mesh and generalisation of the result to the full original 3D mesh of figure\ [@fig:valve-mesh1]
:::::

![An example case where the SCLs are located around the junction between stainless-steel valves and carbon steel pipes at both sides of the material interface in the vertical plane both at the top and at the bottom of the pipe.](valve-scls1-n.png){#fig:valve width=65%}

Note that there is no need to have a one-to-one correspondence between the elements from the reduced mesh with the elements from the original one. Actually, the reduced mesh contains first-order elements whilst the former has second-order elements. Also the grid density is different. Nevertheless, the finite-element solver [Fino](https://www.seamplex.com/fino) used to solve both the heat and the mechanical problems, allows to read functions of space and time defined over one mesh and continuously evaluate and use them into another one even if the two grids have different elements, orders or even dimensions. In effect, in the system from\ [@fig:real-life] the material interface is between a orifice plate made in stainless steel that is welded to a carbon-steel pipe ([@fig:real-mesh2]). The thermal problem can be modelled using a two-dimensional axi-symmetric grid [@fig:real-temp] and then generalised to the full three-dimensional mesh using the algebraic manipulation capabilities provided by [Fino](https://www.seamplex.com/fino) (actually by [wasora](https://www.seamplex.com/wasora)) as shown in\ [@fig:real-gen].
As an example, let us consider the system depicted in\ [@fig:valve] where there is a stainless-carbon steel interface at the discharge of the valves. Instead of solving the transient heat-conduction problem with the internal temperature of the pipes equal to the temperature of the water in the reference transient condition of the power plant and an external condition of natural convection to the ambient temperature in the whole mesh, a reduced model consisting of half of one of the two valves and a small length of the pipes at both the valve inlet and outlet is used. Once the temperature distribution\ $\hat{T}(\mathbf{x},t)$ for each time is obtained in the reduced mesh ([@fig:valve-temp], which has the origin at the centre of the valve), the actual temperature distribution\ $T(\mathbf{x},t)$ is computed by an algebraic generalisation of $\hat{T}(\mathbf{x},t)$ in the full coordinate system. As stated above, those locations which are not covered by the reduced model are generalised with a time-dependent uniform temperature which is the average of the inner and outer temperatures at the inlet and outlet of the reduced mesh.

dnl 33410 10-12D-24
![Reduced mesh around a valve refined around the interface where the transient heat conduction problem is solved.](valve-temp.png){#fig:valve-temp width=80%}

::::: {#fig:real}
![Full 3D mesh, carbon steel is magenta, stainless steel is green.](real-mesh2.png){#fig:real-mesh2 width=70%}

![Temperature distribution for a certain time within the transient computed on a reduced two-dimensional axi-symmetric mesh modelling half the orifice plate and a length of the carbon pipe.](real-temp.png){#fig:real-temp width=100%}
Note that there is no need to have a one-to-one correspondence between the elements from the reduced mesh with the elements from the original one. Actually, the reduced mesh contains first-order elements whilst the former has second-order elements. Also the grid density is different. Nevertheless, the finite-element solver [Fino](https://www.seamplex.com/fino) used to solve both the heat and the mechanical problems, allows to read functions of space and time defined over one mesh and continuously evaluate and use them into another one even if the two grids have different elements, orders or even dimensions. In effect, in the system from\ [@fig:real-life] the material interface is between a orifice plate made in stainless steel that is welded to a carbon-steel pipe ([@fig:real]). The thermal problem can be modelled using a two-dimensional axi-symmetric grid and then generalised to the full three-dimensional mesh using the algebraic manipulation capabilities provided by [Fino](https://www.seamplex.com/fino) (actually by [wasora](https://www.seamplex.com/wasora)) as shown in\ [@fig:real].

![Generalisation of the temperature to the full three-dimensional grid.](real-gen.png){#fig:real-gen width=75%}
dnl ![Full 3D mesh, carbon steel is magenta, stainless steel is green.](real-mesh2.png){#fig:real-mesh2 width=70%}
dnl ![Temperature distribution for a certain time within the transient computed on a reduced two-dimensional axi-symmetric mesh modelling half the orifice plate and a length of the carbon pipe.](real-temp.png){#fig:real-temp width=34%}

The material interface in the system from [@fig:real-life] is configured by an orifice plate made of stainless steel welded to a carbon-steel pipe
:::::
![Temperature distribution for a certain time within the transient computed on a reduced two-dimensional axi-symmetric mesh modelling half the orifice plate and a length of the carbon pipe and Generalisation of the temperature to the full three-dimensional grid.](real-gen.png){#fig:real width=85%}


## Seismic loads {#sec:seismic}
### Seismic loads {#sec:seismic}

Before considering the actual mechanical problem that will give us the stress tensor at the SCLs, and besides needing to solve the transient thermal problem to get the temperature distributions, we need to address the loads that arise due to a postulated earthquake during a certain part of the operational transients. The full computation of a mechanical transient problem using the earthquake time-dependent displacements is off the table for two reasons. First, because again the computation would take more time than we might have to deliver the report. And secondly and more importantly, because civil engineers do not compute earthquakes in the time domain but in the frequency domain using the [response spectrum method](https://en.wikipedia.org/wiki/Response_spectrum). Time to revisit our [Fourier transform](https://en.wikipedia.org/wiki/Fourier_transform) exercises from undergraduate maths courses.
Before considering the actual mechanical problem that will give us the stress tensor at the SCLs, and besides needing to solve the transient thermal problem to get the temperature distributions, we need to address the loads that arise due to a postulated earthquake during a certain part of the operational transients. The full computation of a mechanical transient problem using the earthquake time-dependent displacements is off the table for two reasons. First, because again the computation would take more time than we might have to deliver the report. And secondly and more importantly, because civil engineers do not compute earthquakes in the time domain but in the frequency domain using the [response spectrum method](https://en.wikipedia.org/wiki/Response_spectrum). Time to revisit our [Fourier transform](https://en.wikipedia.org/wiki/Fourier_transform) exercises from undergraduate math courses.

### Earthquake spectra
#### Earthquake spectra

Every nuclear power plant is designed to withstand earthquakes. Of course, not all plants need the same level of reinforcements. Those built in large quiet plains will be, seismically speaking, cheaper than those located in geologically active zones. Keep in mind that all the 54 Japanese nuclear power plants did structurally resist the 2011 earthquake, and all of the reactors were safely shut down. What actually happened in [Fukushima](http://www.world-nuclear.org/information-library/safety-and-security/safety-of-plants/fukushima-accident.aspx) is that one hour after the main shake, a 14-metre tsunami splashed on the coast, jumping over the 9-metre defences and flooding the emergency [Diesel generators](https://en.wikipedia.org/wiki/Diesel_generator) that provided power to the pumps in charge of removing the remaining [decay power](https://en.wikipedia.org/wiki/Decay_heat) from the already-stopped [reactor core](https://en.wikipedia.org/wiki/Nuclear_reactor_core).

Back to our case study, the point is that each site where nuclear power plants are built must have a geological study where a postulated design-basis earthquake is to be defined. In other words, a theoretical earthquake which the plant ought to withstand needs to be specified. How? By giving a set of three spectra (one for each coordinate direction) giving acceleration as a function of the frequency for each level of the building. That is to say, once the earthquake hits the power plant, depending on soil-structure interactions the energy will shake the building foundations in a way that depends on the characteristics of the earthquake, the soil and the concrete structure. Afterwards, the way the oscillations travel upward and shake each of the mechanical components erected in each floor level depends on the design of the civil structure in a way which is fully determined by the floor response spectra like the ones depicted in\ [@fig:spectrum].

![A sample spectrum for a certain floor level of a certain nuclear power plant.](spectrum.svg){#fig:spectrum width=90%}
![A sample spectrum for a certain floor level of a certain nuclear power plant.](spectrum.svg){#fig:spectrum width=70%}

### Natural frequencies
#### Natural frequencies

As the earthquake excites some frequencies more than others, it is mandatory to know which are the natural frequencies and modes of oscillations of our piping system. Mathematically, this requires the computation of an eigenvalue problem. Simply stated, we need to find all the non-trivial solutions of the equation

@@ -514,7 +427,7 @@ The equivalent accelerations for the piping section of [@fig:modes] for the spec
The ASME code says that these accelerations (depicted in [@fig:acceleration]) ought to be applied twice: once with the original sign and once with all the elements with the opposite sign. Each application should last two seconds.


## Linearity (not yet linearisation) {#sec:linearity}
### Linearity (not yet linearisation) {#sec:linearity}

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 during a couple of seconds a statically-equivalent distributed load arising 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 to exploit the problem’s linearity and compute all the effects separately and then sum them up to obtain the whole combination. We may thus compute just the stresses due to the seismic loads and then add these stresses to the stresses at any time 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 result of the sum of the cases, right? Not always.

@@ -536,9 +449,10 @@ Now we are going to create and compare three load cases:

The loads in each cases are applied to the three remaining faces, namely “right” ($x>0$), “back” ($y>0$) and “top,” ($z>0$). Their magnitude in Newtons are:

````{=latex}
::::::{=latex}
\rowcolors{2}{black!10}{black!0}
````
::::::

| | | “right” | | | “back” | | | “top” | |
| ------ |:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|
| | $F_x$ | $F_y$ | $F_z$ | $F_x$ | $F_y$ | $F_z$ | $F_x$ | $F_y$ | $F_z$ |
@@ -546,86 +460,10 @@ The loads in each cases are applied to the three remaining faces, namely “righ
| Case B | 0 | +15 | -15 | +25 | 0 | -5 | -15 | +25 | 0 |
| Case C | +10 | +15 | -15 | +25 | +20 | -5 | -15 | +25 | +30 |

divert(-1)

````{=latex}
\begin{center}
\rowcolors{2}{black!10}{black!0}
\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 & 0 \\
Case C, combination & +10 & +15 & -15 & +25 & +20 & -5 & -15 & +25 & +30 \\
\end{tabular}
\end{center}
````

````{=html}
<table class="table table-responsive w-100">
<tr>
<th></th>
<th colspan="3">face “right” (<span class="math inline">x&gt;0</span>)</th>
<th colspan="3">face “back” (<span class="math inline">y&gt;0</span>)</th>
<th colspan="3">face “top” (<span class="math inline">z&gt;0</span>)</th>
</tr>
<tr>
<td></td>
<td><span class="math inline">F_x</span></td>
<td><span class="math inline">F_y</span></td>
<td><span class="math inline">F_z</span></td>
<td><span class="math inline">F_x</span></td>
<td><span class="math inline">F_y</span></td>
<td><span class="math inline">F_z</span></td>
<td><span class="math inline">F_x</span></td>
<td><span class="math inline">F_y</span></td>
<td><span class="math inline">F_z</span></td>
</tr>
<tr>
<td>Case A, pure normal</td>
<td>+10</td><td>0</td><td>0</td><td>0</td><td>+20</td><td>0</td><td>0</td><td>0</td><td>+30</td>
</tr>
<tr>
<td>Case B, pure shear</td>
<td>0</td><td>+15</td><td>-15</td><td>+25</td><td>0</td><td>-5</td><td>-15</td><td>+25</td><td>0</td>
</tr>
<tr>
<td>Case C, combination</td>
<td>+10</td><td>+15</td><td>-15</td><td>+25</td><td>+20<td>-5</td><td>-15</td><td>+25</td><td>+30</td>
</tr>
</table>
````
divert(0)


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 1\ mm$^2$, the usual sorting leads to

divert(-1)
$$
\begin{align*}
\sigma_{1A} &= 30~\text{MPa} \\
\sigma_{2A} &= 20~\text{MPa} \\
\sigma_{3A} &= 10~\text{MPa} \\
\end{align*}
$$
divert(0)

$$
\sigma_{1A} = 30~\text{MPa}
$$
@@ -707,16 +545,16 @@ then the sums (in plural because there are three eigenvalues) of their eigenvalu
The moral of this fable is that if we have a case that is the combination of two other simpler cases (say one with only surface loads and one with only volumetric loads), in general we cannot just add the principal stresses (or Von Mises) of two cases and expect to obtain a correct answer. We have to solve the full case again (both the surface and the volumetric loads at the same time). In case we are stubborn enough and still want to stick to solving the cases separately, there is a trick we can resort to. Instead of summing principal stresses, what we can do is to sum either displacements or the individual stress components, which are fully linear. So we might pre-deform (or pre-stress) case B with the results from case A and then expect the FEM program to obtain the correct stresses for the combined case. However, this scheme is actually far more complex than just solving the combined case in a single run and it also needs an educated guess and/or trial and error to know at what time the pre-deformation or pre-stressing should be applied to take into account the seismic load.


## ASME stress linearisation (not linearity!)
### ASME stress linearisation (not linearity!)

After discussing linearity, let us now dig into linearisation. The name is similar but these two animals are very different beasts. We said in\ [@sec:case] that the ASME Boiler and Pressure Vessel Code was born long before modern finite-elements methods were developed and of course being massively available for general engineering analysis (democratised?). Yet the code provides a comprehensive, sound and, more importantly, a widely and commonly-accepted body of knowledge as for the regulatory authorities to require its enforcement to nuclear plant owners. One of the main issues of the ASME code refers to what is known as “membrane” and “bending” stresses. These are defined in [section\ VIII](https://en.wikipedia.org/wiki/ASME_Boiler_and_Pressure_Vessel_Code#ASME_BPVC_Section_VIII_-_Rules_for_Construction_of_Pressure_Vessels) annex 5-A, although they are widely used in other sections, particularly [section\ III](https://en.wikipedia.org/wiki/ASME_Boiler_and_Pressure_Vessel_Code#ASME_BPVC_Section_III_-_Rules_for_Construction_of_Nuclear_Facility_Components). Briefly, they give the zeroth-order (membrane) and first-order (bending) [moments](https://en.wikipedia.org/wiki/Moment_(mathematics)) (in the statistical sense) of the stress distribution along a so-called Stress Classification Line or SCL, which should be chosen depending on the type of problem under analysis.

The computation of these membrane and bending stresses are called [“stress linearisation”](https://www.ramsay-maunder.co.uk/knowledge-base/technical-notes/stress-linearisation-for-practising-engineers/) because (I am guessing) it is like computing the [Taylor expansion](https://en.wikipedia.org/wiki/Taylor_series) (or for the case, expansion in [Legendre polynomials](https://en.wikipedia.org/wiki/Legendre_polynomials)) of an arbitrary stress distribution along a line, and retaining the first two terms. That is to say, to obtain a linear approximation. More (optional) mathematical details below.

dnl **figures**

So what about the SCLs? Well, the ASME standard says that they are lines that go through a wall of the pipe (or vessel or pump, which is what the ASME code is for) from the inside to the outside and ought to be normal to the iso-stress curves. Stop. Picture yourself a stress field, draw the iso-stress curves (those would be the lines that have the same colour in your picture) and then imagine a set of lines that travel in a perpendicular direction to them. Finally, choose the one that seems the prettiest (which most of the time is the one that seems the easiest). There you go! You have an SCL. But there is a catch. So far, we have referred to a generic concept of “stress.” Which of the several stresses out there should you picture? One of the three normals, the three shear, Von\ Mises, Tresca? Well, actually you will have to imagine tensors instead of scalars. And there might not be such a thing as “iso-stress” curves, let alone normal directions. So pick any radial straight line through the pipe wall at a location that seems relevant and now you are done. In our case study, there will be a few different locations around the material interfaces where high stresses due to differential thermal expansion are expected to occur.

divert(-1)
\medskip

Now the optional (but recommended) mathematical details. Accoding to ASME\ VIII Div.\ 2 Annex\ 5-A, the expression for computing the $i$-$j$-th element of the membrane tensor is
@@ -764,9 +602,9 @@ $$

No need to know or even understand these integrals which for sure are not introduced to students in regular college courses. But it would be good to, as linearisation is a cornerstone subject for any serious mechanical analysis of pressurised components following the code. So get a copy of ASME sec.\ VIII div.\ 2 annex\ 5-A and then search online for for “stress linearisation” (or “linearization”).

divert(0)


# The infinite pipe revisited after college {#sec:infinite-pipe-fem}
## The infinite pipe revisited after college {#sec:infinite-pipe-fem}

Let us now make a (tiny) step from the general and almost philosophical subject from the last section down to the particular case study, and reconsider the infinite pressurised pipe once again. It is time to solve the problem with a computer using finite elements and to obtain some funny coloured pictures instead of just equations.

@@ -791,17 +629,18 @@ Two of the hundreds of different ways the infinite pressurised pipe can be solve
:::::


You can get both the exponential nature of each added bullet and how easily we can add new further choices to solve a FEM problem. And each of these choices will reveal you something about the nature of either the mechanical problem or the numerical solution. It is not possible to teach any possible lesson from every outcome in college, so you will have to learn them by yourself getting your hands at them. I have already tried to address the particular case of the infinite pipe in a [recent report](https://www.seamplex.com/fino/doc/pipe-linearized/)^[<https://www.seamplex.com/fino/doc/pipe-linearized/>] that is worth reading before carrying on with this article. The conclusions of the report are:
You can get both the exponential nature of each added bullet and how easily we can add new further choices to solve a FEM problem. And each of these choices will reveal you something about the nature of either the mechanical problem or the numerical solution. It is not possible to teach any possible lesson from every outcome in college, so you will have to learn them by yourself getting your hands at them. I have already tried to address the particular case of the infinite pipe in a [recent report](https://www.seamplex.com/fino/doc/pipe-linearized/)^[<https://www.seamplex.com/fino/doc/pipe-linearized/>] that is worth reading before carrying on with this article.
The main conclusions of the report are:

* Engineering problems ought not to be solved using black-boxes (i.e. privative software whose source code is not freely available)---more on the subject below in\ [@sec:two-materials].
* The pressurised infinite pipe has only one independent variable (the radius $r$) and one primary dependent variable (the radial displacement $u_r$).
* The problem has analytical solution for the radial displacement\ $u_r$ and the radial\ $\sigma_r$, tangential\ $\sigma_\theta$ and axial\ $\sigma_z$ stresses.
* There are no shear stresses, so these three stresses are also the principal stresses.
* Analytical expressions for the membrane and membrane plus bending stresses along any radial SCL can be obtained.
* The spatial domain can be discretised using linear or higher-order elements. In particular first and second-order elements have been used in the report.
dnl * Engineering problems ought not to be solved using black-boxes (i.e. privative software whose source code is not freely available)---more on the subject below in\ [@sec:two-materials].
dnl * The pressurised infinite pipe has only one independent variable (the radius $r$) and one primary dependent variable (the radial displacement $u_r$).
dnl * The problem has analytical solution for the radial displacement\ $u_r$ and the radial\ $\sigma_r$, tangential\ $\sigma_\theta$ and axial\ $\sigma_z$ stresses.
dnl * There are no shear stresses, so these three stresses are also the principal stresses.
dnl * Analytical expressions for the membrane and membrane plus bending stresses along any radial SCL can be obtained.
dnl * The spatial domain can be discretised using linear or higher-order elements. In particular first and second-order elements have been used in the report.
* For the same number of elements, second-order grids need more nodes than linear ones, although they can better represent curved geometries.
* The discretised problem size depends on the number of nodes and not on the number of elements---more on the subject below in\ [@sec:elements-nodes].
* The finite-element results for the displacements are similar to the analytical solution, with second-order grids giving better results for the same number of elements (this is expected as they involved far more nodes).
dnl * The finite-element results for the displacements are similar to the analytical solution, with second-order grids giving better results for the same number of elements (this is expected as they involved far more nodes).
* The three stress distributions computed with the finite-element give far more reasonable results for the second-order case than for the first-order grid. This is qualitatively explained by the fact that first-order tetrahedra have uniform derivatives and such the elements located in both the external and external faces represent the stresses not at the actual faces but at the barycentre of the elements.
* Membrane stresses converge well for both the first and second-order cases because they represent a zeroth-order moment of the stress distribution and the excess and defect errors committed by the internal and external elements approximately cancel out.
* Membrane plus bending stresses converge very poorly with linear elements because the excess and defect errors do not cancel out because it is a first-order moment of the stress distribution.
@@ -809,16 +648,15 @@ You can get both the exponential nature of each added bullet and how easily we c
* The error with respect to the analytical solutions as a function of the CPU time needed to compute the membrane stress is similar for both first and second-order grids. But for the computation of the membrane plus bending stress ([@fig:error-MB-vs-cpu]), first-order grids give very poor results compared to second-order grids for the same CPU time.

::::: {#fig:error-vs-cpu}
![Membrane $\text{M}$](error-M-vs-cpu.svg){#fig:error-M-vs-cpu width=90%}

![Membrane plus bending $\text{MB}$](error-MB-vs-cpu.svg){#fig:error-MB-vs-cpu width=90%}
![Membrane $\text{M}$](error-M-vs-cpu-small.svg){#fig:error-M-vs-cpu width=49%}
![Membrane plus bending $\text{MB}$](error-MB-vs-cpu-small.svg){#fig:error-MB-vs-cpu width=49%}

Error in the computation of the linearised stresses vs. CPU time needed to solve the infinite pipe problem using the finite element method
:::::

An additional note should be added. The FEM solution, which not only gives the nodal displacements but also a method to interpolate these values inside the elements, does not fully satisfy the original equilibrium equations at every point (i.e. the strong formulation). It is an approximation to the solution of the [weak formulation](https://en.wikipedia.org/wiki/Weak_formulation) that is close (measured in the vector space spanned by the [shape functions](https://www.quora.com/What-is-a-shape-function-in-FEM)) to the real solution. Mechanically, this means that the FEM solution leads only to nodal equilibrium but not pointwise equilibrium.

## Elements, nodes and CPU {#sec:elements-nodes}
### Elements, nodes and CPU {#sec:elements-nodes}

The last two bullets above lead to an issue that has come many times when discussing the issue of convergence with respect to the mesh size with other colleagues. There apparently exists a common misunderstanding that the number of elements is the main parameter that defines how complex a FEM model is. This is strange, because even in college we are taught that the most important parameter is the _size_ of the stiffness matrix, which is three times (for 3D problems with the [displacement-based formulation](http://web.mit.edu/kjb/www/Books/FEP_2nd_Edition_4th_Printing.pdf) formulation) the number of _nodes_.

@@ -837,17 +675,17 @@ To fix ideas, let us stick to a linear elastic FEM problem. The CPU time needed
3. solving the equations to obtain the displacements, and
4. computing the stress from the displacements.

### Meshing {#sec:meshing}
#### Meshing {#sec:meshing}

The effort needed to compute a discretisation of a continuous domain depends on the meshing algorithm. But nearly all meshers first put nodes on the edges (1D), then on the surfaces (2D) and finally on the volumes (3D). Afterwards, they join the nodes to create the elements. Depending on the topology (i.e. tetrahedra, hexahedra, pyramids, etc) and the order (i.e. linear, quadratic, etc.) this last step will vary, but the main driver here is the number of nodes. Try measuring the time needed to obtain grids of different sizes and kinds with your mesher.

### Building {#sec:building}
#### Building {#sec:building}

The [stiffness matrix]([stiffness matrix](https://en.wikipedia.org/wiki/Stiffness_matrix)) is a square matrix that has\ $NG$ rows and\ $NG$ columns where $N$ is the number of nodes and $G$ is the number of degrees of freedom per node, which for three-dimensional problems is $G=3$. But even though FEM problems have to build a $NG\times NG$ matrix, they usually sweep through elements rather than through nodes, and then scatter the elements of the elemental matrices to the global stiffness matrix. This is called the assembly of the matrix. So the effort needed here depends again on how the solver is programmed, but it is a combination of the number of elements and the number of nodes.

For a fixed number of nodes, first-order grids have far more elements than second-order grids because in the first case each node has to be a vertex while in the latter half will be vertexes and half will be located at the edges (think!). So the sweep is larger for linear grids. But the effort needed to integrate quadratic shape functions is greater than for the linear case, so these two effects almost cancel out.

### Solving {#sec:solving}
#### Solving {#sec:solving}

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\ $\mathbf{b}$ containing the loads (both volumetric and the ones at the surfaces from the boundary conditions):

@@ -869,7 +707,7 @@ Structure of the stiffness matrices for the same FEM problem with 10k nodes. Red

In a similar way, different types of elements will give rise to different sparsity patterns which change the effort needed to solve the problem. In any case, the base parameter that controls the problem size and thus provides a basic indicator of the level of difficulty the problem poses is the number of nodes. Again, not the number of elements, as the solver does not even know if the matrix comes from FEM, FVM or FDM.

### Stress computation {#sec:stress-computation}
#### Stress computation {#sec:stress-computation}

In the [displacement-based formulation](http://web.mit.edu/kjb/www/Books/FEP_2nd_Edition_4th_Printing.pdf), the solver finds the displacements\ $\mathbf{u}(\mathbf{x})$ that satisfy [@eq:kub], which are the principal unknowns. But from\ [@sec:tensor] we know that we actually have solved the problem after we have the stress tensors at every location\ $\mathbf{x}$, which are the secondary unknowns. So the FEM program has to compute the stresses out of the displacements. It first computes the strain tensor, which is composed of the nine partial derivatives of the three displacements with respect to the three coordinates. Then it computes the stress tensor (atready introduced in\ [@sec:tensor]) using the materials’ [strain-stress constitutive equations](https://en.wikipedia.org/wiki/Constitutive_equation#Stress_and_strain) which involve the Young Modulus\ $E$, the Poisson ratio\ $\nu$ and the spatial derivatives of the displacements\ $\mathbf{u}=[u,v,w]$. This sounds easy, as we (well, the solver) knows what the shape functions are for each element and then it is a matter of computing nine derivatives and multiplying by something involving\ $E$ and\ $\nu$. Yes, but there is a catch. As the displacements\ $u$, $v$ and\ $w$ are computed at the nodes, we would like to also have the stresses at the nodes. However,

@@ -895,11 +733,11 @@ Detailed mathematics show that the location where the derivatives of the interpo
In any case, this step takes a non-negligible amount of time. The most-common approach, i.e. the node-averaging method is driven mainly by the number of nodes of course. So all-in-all, these are the reasons to use the number of nodes instead of the numbers of elements as a basic parameter to measure the complexity of a FEM problem.


# Adding complexity: the truth is out there
## Adding complexity: the truth is out there

Let us review some issues that appear when solving our case study and that might not have been thoroughly addressed back during our college days.

## Two (or more) materials {#sec:two-materials}
### Two (or more) materials {#sec:two-materials}

The main issue with fatigue in nuclear piping during operational transients is that at the welds between two materials with different thermal expansion coefficients there can appear potentially-high stresses during temperature changes. If these transients are repeated cyclically, fatigue may occur. We already have risen a warning flag about stresses at material interfaces in\ [@sec:two-materials]. Besides all the open questions about computing stresses at nodes, this case also adds the fact that the material properties (say the Young Modulus\ $E$) is different in the elements that are at each side of the interface.

@@ -919,20 +757,9 @@ $$

Faced with the problem of computing the stress\ $\sigma$ at one node shared by many elements, we might:

1. compute the (weighted?) averages\ $\langle E \rangle$ and $\langle \epsilon \rangle$ and then compute the stress as

$$ \langle \sigma \rangle = \langle E \rangle \cdot \langle \epsilon \rangle$$

This would be like having a meta-material at the interface with average properties, or

2. compute the stress as the (weighted?) average of the product $E \cdot \epsilon$ in each node

$$ \langle \sigma \rangle = \langle E \cdot \epsilon \rangle$$

This would be like forcing a non-differentiable function to behave smoothly, or

1. compute the (weighted?) averages\ $\langle E \rangle$ and $\langle \epsilon \rangle$ and then compute the stress as\ $\langle \sigma \rangle = \langle E \rangle \cdot \langle \epsilon \rangle$. This would be like having a meta-material at the interface with average properties, or
2. compute the stress as the (weighted?) average of the product $E \cdot \epsilon$ in each node\ $\langle \sigma \rangle = \langle E \cdot \epsilon \rangle$. This would be like forcing a non-differentiable function to behave smoothly, or
3. internally duplicate the nodes at the interface and compute one stress for each material. This would result in a stress distribution which is not a real function because the same location\ $\mathbf{x}$ will be associated to more than one stress value, or

4. duplicate the surface elements at the interfaces and solve the problem using a contact formulation. This would render the problem non-linear and add the complexity of having to find appropriate penalty coefficients.

There might be other choices as well. Do you know what your favourite FEM program does? Now follow up with these questions:
@@ -956,9 +783,10 @@ In effect, a couple of years ago Angus Ramsay noted [a weird behaviour](https://

> Do you trust your favourite FEM program?

Back to the two-material problem, all the discussion above in\ [@sec:two-materials] about non-continuous derivatives applies to a sharp abrupt interface. In the study case the junctions are welded so there is a [heat-affected zone](https://en.wikipedia.org/wiki/Heat-affected_zone) with changes in the material micro structure. Therefore, there exists a smooth transition from the mechanical properties of one material to the other one in a way that is very hard to predict and to model. In principle, the assumption of a sharp interface is conservative (in the sense of\ [@sec:kinds]). There cannot be an SCL exactly on a material interface so there should be at least two SCLs, one at each side of the junctions of\ [@fig:weldolet-scls;@fig:valve-scls1] illustrate. The actual distance would have to be determined first as an educated guess, then via trial and error and finally in accordance with the regulator.
Back to the two-material problem, all the discussion above in\ [@sec:two-materials] about non-continuous derivatives applies to a sharp abrupt interface. In the study case the junctions are welded so there is a [heat-affected zone](https://en.wikipedia.org/wiki/Heat-affected_zone) with changes in the material micro structure. Therefore, there exists a smooth transition from the mechanical properties of one material to the other one in a way that is very hard to predict and to model. In principle, the assumption of a sharp interface is conservative (in the sense of\ [@sec:kinds]). There cannot be an SCL exactly on a material interface so there should be at least two SCLs, one at each side of the junctions as\ [@fig:weldolet-scls] illustrates. The actual distance would have to be determined first as an educated guess, then via trial and error and finally in accordance with the regulator.

## A parametric tee {#sec:parametric}
divert(-1)
### A parametric tee {#sec:parametric}

Time for another experiment. We know (more or less) what to expect from an infinite pressurised pipe from\ [@sec:infinite-pipe] and [@sec:infinite-pipe-fem]. What if we added a branch to such pipe? Even more, what if we successively varied the diameter of the branch to see what happens? This is called parametric analysis, and sooner or later (if not now) you will find yourself performing this kind of computations more often than any other one.

@@ -1036,8 +864,9 @@ Do you now see the added value of training throw-ins with watermelons? We might
Most of the time at college we would barely do what is needed to be approved in one course and move on to the next one. If you have the time and consider a career related to finite-element analysis, please do not. Clone the repository^[<https://bitbucket.org/seamplex/tee>] with the input files for [Fino](https://www.seamplex.com/fino) and start playing. If you are stuck, do not hesitate asking for help in [wasora’s mailing list](https://www.seamplex.com/lists.html).

One further detail: it is always a sane check to try to explain the numerical results based on physical reasoning (i.e. “with your fingers”) as we did two paragraphs above. Most of the time you will be solving problems whilst already knowing what the result would (or ought to) be.
divert(0)

## Bake, shake and break {#sec:break}
### Bake, shake and break {#sec:break}

A fellow mechanical engineer who went to the same high school I did, who went to the same engineering school I did and who worked at the same company I did, but who earned a PhD in Norway once told me two interesting things about finite-elements graduate courses. First, that in Trondheim the classes were taught by faculty from the the mathematics department rather than from the mechanical engineering department. It made complete sense to me, as I always have thought finite elements mainly as a maths subject. And even though engineers might know some maths, it is nothing compared to actual mathematicians. Secondly, that they called the thermal, natural oscillations and elastic problems as the rhyming trio “bake, shake and break” (they also had “wake” for fluids, but that is a different story). These are just the three problems listed in section\ [@sec:piping-nuclear] that we need to solve in our nuclear power plant.

@@ -1082,20 +911,12 @@ $$ K(\mathbf{x}) \cdot \mathbf{u}(\mathbf{x}) = \mathbf{b}(\mathbf{x})$$
And this last equation is linear in\ $\mathbf{u}$. In effect, the discretisation step means to integrate over\ $\mathbf{x}$. As\ $K$, $\mathbf{u}$ and\ $\mathbf{b}$ depend only on\ $\mathbf{x}$, 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 (in increasing order of recommendation).


::::: {#fig:case-cad}
![General view. Carbon steel is grey and stainless steel is green](case-cad1.png){#fig:case-cad1 width=75%}
::::: {#fig:case}
![General view. Carbon steel is grey and stainless steel is green.](case-cad1.png){#fig:case-cad1 width=75%}

![Detail of the orifice plate](case-cad2.png){#fig:case-cad2 width=95%}
![Detail of the mesh refinement around the interface.](case-mech-mesh2.png){#fig:case-mesh2 width=100%}

Three-dimensional\ CAD model of a section of pipes system between appropriate supports
:::::

::::: {#fig:case-mesh}
![General view](case-mech-mesh1.png){#fig:case-mesh1 width=85%}

![Detail of the refinement around the interface](case-mech-mesh2.png){#fig:case-mesh2 width=100%}

Unstructured grid for the mechanical analysis. Volumetric elements (i.e. tetrahedra) corresponding to carbon steel are magenta and to stainless steel are green. Surface elements (i.e. triangles) corresponding to the internal pressurised face are yellow
A section of a piping system in a nuclear power plant
:::::

![Location of the 15\ SCLs](case-scls-n.png){#fig:case-scls width=75%}
@@ -1124,28 +945,15 @@ Temperature distribution for a certain instant of the transient, computed in the
First nine natural modes of oscillation of the piping system subject to the boundary conditions the supports provide
:::::

![Floor response spectrum.](case-spectrum.svg){#fig:case-spectrum width=80%}

::::: {#fig:case-acceleration}
![$a_x$](case-ax.png){width=50%}

![$a_y$](case-ay.png){width=50%}

![$a_z$](case-az.png){width=50%}

The static equivalent accelerations for the spectra of\ [@fig:case-spectrum] computed using the SRSS method
:::::


\medskip
![Floor response spectrum.](case-spectrum.svg){#fig:case-spectrum width=70%}

To recapitulate, the figures in this section show some partial non-dimensional results of an actual system of a certain nuclear power plant. The main issues to study were the interfaces between a carbon-steel pipe and a stainless-steel orifice plate used to measure the (heavy) water flow through the line. The steps discussed so far include

1. building a CAD model of the piping section under study, which will be the main domain ([@fig:case-cad] or [@fig:real-life;@fig:weldolet-cad;@fig:valve-cad1])
2. creating a mesh for the main domain refining locally around the material interfaces ([@fig:case-mesh] or [@fig:weldolet-mesh;@fig:valve-mesh1;@fig:real-mesh2])
3. defining the number and locations of the SCLs ([@fig:case-scls] or [@fig:weldolet-scls;@fig:valve-scls1;@fig:tee-scls])
4. computing a heat conduction (bake) transient problem with temperatures as a function of time from the operational transient in a simple domain using temperature-dependent thermal conduction coefficients from ASME\ II\ part D ([@fig:case-temp] or [@fig:valve-temp;@fig:real-mesh2])
5. generalising the temperature distribution as a function of time to the general domain ([@fig:case-temp2] or [@fig:valve-gen;@fig:real-gen])
1. building a CAD model of the piping section under study, which will be the main domain ([@fig:case-cad1] or [@fig:real-life;@fig:weldolet-cad;@fig:valve])
2. creating a mesh for the main domain refining locally around the material interfaces ([@fig:case-mesh2] or [@fig:weldolet-mesh;@fig:real])
3. defining the number and locations of the SCLs ([@fig:case-scls] or [@fig:weldolet-scls;@fig:valve])
4. computing a heat conduction (bake) transient problem with temperatures as a function of time from the operational transient in a simple domain using temperature-dependent thermal conduction coefficients from ASME\ II\ part D ([@fig:case-temp] or [@fig:valve-temp])
5. generalising the temperature distribution as a function of time to the general domain ([@fig:case-temp2] or [@fig:real])
6. performing a modal analysis (shake) on the main domain to obtain the main oscillation frequencies and modes ([@fig:case-mode] or [@fig:modes])
7. using the floor response spectra ([@fig:case-spectrum] or [@fig:spectrum]) and the SRSS method to obtain a distributed force statically-equivalent to the earthquake load ([@fig:case-acceleration] or [@fig:acceleration])
8. successively solving the linear elastic problem for different times using the generalised temperature distribution taking into account
@@ -1154,17 +962,21 @@ To recapitulate, the figures in this section show some partial non-dimensional r
c. the instantaneous pressure exerted in the internal faces of the pipes at the time\ $t$ according to the definition of the operational transient
d. the restriction of the degrees of freedom of those faces, lines or points that correspond to mechanical supports located both within and at the ends of the CAD model
e. the earthquake load, which according to ASME should be present only during four seconds of the transient: two seconds with one sign and the other two seconds with the opposite sing. This period should be selected to coincide with the instant of highest mechanical stress (conservative computation)
9. computing the linearised stresses (membrane and membrane plus bending) at the SCLs combining them as
a. Principal 1
b. Principal 2
c. Principal 3
d. Tresca
9. computing the linearised stresses (membrane and membrane plus bending) at the SCLs combining them as Principal\ 1, Principal\ 2, Principal\ 3 and Tresca
10. juxtaposing these linearised stresses for each time of the transient and for each transient so as to obtain a single time-history of stresses including all the operational and/or incidental transients under study, which is what stress-based fatigue assessment needs (recall\ [@sec:fatigue] and go on to\ [@sec:usage]).

A pretty nice list of steps, which definitely I would not have been able to tackle when I was in college. Would you?


# Cumulative usage factors {#sec:usage}
::::: {#fig:case-acceleration}
![$a_x$](case-ax.png){width=33%}
![$a_y$](case-ay.png){width=33%}
![$a_z$](case-az.png){width=33%}

The static equivalent accelerations for the spectra of\ [@fig:case-spectrum] computed using the SRSS method
:::::

## Cumulative usage factors {#sec:usage}

Strictly speaking, finite elements are not needed anymore at this point of the analysis. But even though we are (or want to be) FEM experts, we have to understand that if the objective of a work is to evaluate fatigue (or fracture mechanics or whatever), finite elements are just a mean and not and end. If we just mastered FEM and nothing else, our field of work would be highly reduced. We need to use all of our computational knowledge to perform actually engineering tasks and to be able to tell our bosses and clients whether the pipe would fail or not. This tip is induced in college but it is definitely reinforced afterwards when working with actual clients and bosses.

@@ -1174,7 +986,7 @@ Another comment I would like to add is that I had to learn fatigue practically f

Back and distantly, in\ [@sec:case] we said that people noticed there were some environmental factors that affected the fatigue resistance of materials. The basic ASME approach does not take care of these factors, and it is regarded as fatigue “in air.” We are interested in taking them into account, so we follow the US Nuclear Regulatory Commission guidelines to evaluate fatigue “in water.”

## In air (ASME’s basic approach) {#sec:in-air}
### In air (ASME’s basic approach) {#sec:in-air}

We already said in\ [@sec:fatigue] that the stress-life fatigue assessment method gives the limit number\ $N$ of cycles that a certain mechanical part can withstand when subject to a certain periodic load of stress amplitude\ $S_\text{alt}$. If the actual number of cycles\ $n$ the load is applied is smaller than the limit\ $N$, then the part is fatigue-resistant. In our case study there is a mixture of several periodic loads, each one expected to occur a certain number of times. ASME’s way to evaluate the resistance is to break up the stress history into partial stress amplitudes\ $S_{\text{alt},j}$ between a “peak” and a “valley” and to compute individual usage factors\ $U_j$ for the\ $j$-th amplitude (which does not need to coincide with one of the \ $k$ transient loads) as

@@ -1184,16 +996,16 @@ The overall cumulative usage factor is then the algebraic sum of the partial con

$$\text{CUF} = U_1 + U_2 + \dots + U_j + \dots$$

If\ $\text{CUF} < 1$, then the part under analysis can withstand the proposed cyclic operation.
When\ $\text{CUF} < 1$, then the part under analysis can withstand the proposed cyclic operation.

![A low-alloy steel vessel nozzle (blue) welded to a stainless steel pipe (grey) for the “EAF Sample Problem 2-Rev.\ 2 (10/21/2011)”](axi-inches-3d.png){#fig:axi-inches-3d width=80%}
![A low-alloy steel vessel nozzle (blue) welded to a stainless steel pipe (grey) for the “EAF Sample Problem 2-Rev.\ 2 (10/21/2011)”](axi-inches-3d.png){#fig:axi-inches-3d width=60%}

If the extrema of the partial stress amplitude correspond to different transients, then the following note in ASME III’s NB-3224(5) should be followed:
Now, if the extrema of the partial stress amplitude correspond to different transients, then the following note in ASME III’s NB-3224(5) should be followed:

> In determining $n_1$, $n_2$, $n_3$, $\dots$, $n_j$ consideration shall be given to the superposition of cycles of various origins which produce a total stress difference range greater than the stress difference ranges of the individual cycles. For example, if one type of stress cycle produces 1,000 cycles of a stress difference variation from zero to +60,000\ psi and another type of stress cycle produces 10,000 cycles of a stress difference variation from zero to −50,000\ psi, the two types of cycle to be considered are defined by the following parameters:
>
> (a) for type 1 cycle, $n_1 =$ 1,000 and $S_{\text{alt},1} = (60,000 + 50,000)/2 = 55,000$\ psi;
> (b) for type 2 cycle, $n_2 =$ 9,000 and $S_{\text{alt},2} = (50,000 + 0)/2 = 25,000$\ psi.
> (a) for type 1 cycle, $n_1 =$ 1,000 and $S_{\text{alt},1} = (60,000 + 50,000)/2;
> (b) for type 2 cycle, $n_2 =$ 9,000 and $S_{\text{alt},2} = (50,000 + 0)/2.

This cryptic paragraph can be better explained by using a clearer example. To avoid using actual sensitive data from a real power plant, let us use the same test case used by both the [US Nuclear Regulatory Commission](https://en.wikipedia.org/wiki/Nuclear_Regulatory_Commission) (in its report NUREG/CR-6909) and the [Electric Power Institute](https://en.wikipedia.org/wiki/Electric_Power_Research_Institute) (report 1025823) called “EAF (Environmentally-Assisted Fatigue) Sample Problem 2-Rev.\ 2 (10/21/2011)”.

@@ -1209,11 +1021,11 @@ It consists of a typical vessel (NB-3200) nozzle with attached piping (NB-3600)
A design-basis earthquake was assumed to occur briefly during one second (sic) at around\ $t=3470$\ seconds, and it is assigned a number of cycles\ $n_e=5$. The detailed stress history for one of the SCLs including both the principal and lineariased stresses, which are already offset following NB-3216.2 so as to have a maximum stress equal to zero, can be found as an appendix in NRC's NUREG/CR-6909 report, or in the repository with the scripts I prepared for you to play with this problem.^[<https://bitbucket.org/seamplex/cufen>]

::::: {#fig:nureg}
![Full time range of the juxtaposed transients spanning $\approx$ 36,000 seconds](nureg1.svg){#fig:nureg1 width=75%}
![Full time range of the juxtaposed transients spanning $\approx$ 36,000 seconds](nureg1.svg){#fig:nureg1 width=70%}

![Detail of transients showing the ids of some extrema](nureg2.svg){#fig:nureg2 width=75%}
![Detail of transients showing the ids of some extrema](nureg2.svg){#fig:nureg2 width=70%}

![Detail of the earthquake (it does not follow the ASME two-second rule)](nureg3.svg){#fig:nureg3 width=75%}
![Detail of the earthquake (it does not follow the ASME two-second rule)](nureg3.svg){#fig:nureg3 width=70%}

Time history of the linearised stress\ $\text{MB}_{31}$ corresponding to the example problem from NRC and EPRI. The indexes\ $i$ of the extrema are shown in green (minimums) and red (maximums)
:::::
@@ -1236,38 +1048,30 @@ We now need to comply with ASME’s obscure note about the number of cycles to a

![Results reproduced by the author](cuf-seamplex.png){#fig:cuf-seamplex width=100%}

Tables of individual usage factors for the NRC/EPRI “EAF Sample Problem 2-Rev.\ 2 (10/21/2011).” One table is taken from a document issued by almost-a-billion-dollar-year-budget government agency from the most powerful country in the world and the other one is from a third-world engineering startup
Tables of individual usage factors for the NRC/EPRI “EAF Sample Problem 2-Rev.\ 2 (10/21/2011).” One table is taken from a document issued by almost-a-billion-dollar-year-budget government agency from the most powerful country in the world and the other one is from a third-world engineering startup. Guess which is which.
:::::

Why all these details? Not because I want to teach you how to perform fatigue evaluations just reading this section without resorting to ASME, fatigue books and even other colleagues. It is to show that even though these computation can be made “by hand” (i.e. using a calculator or, God forbids, a spreadsheet) when having to evaluate a few SCLs within several piping systems it is far (and I mean really far) better to automatise all these steps by writing a set of scripts. Not only will the time needed to process the information be reduced, but also the introduction of human errors will be minimised and repeatability of results will be assured---especially if working under a [distributed version control](https://en.wikipedia.org/wiki/Distributed_version_control) system such as [Git](https://en.wikipedia.org/wiki/Git). This is true in general, so here is another tip: learn to write scripts to post-process your FEM results (you will need to use a script-friendly FEM program) and you will gain considerable margins regarding time and quality.

## In water (NRC’s extension) {#sec:in-water}
### In water (NRC’s extension) {#sec:in-water}

The fatigue curves and ASME’s (both\ III and VIII) methodology to analyse cyclic operations assume the parts under study are in contact with air, which is not the case of nuclear reactor pipes. Instead of building a brand new body of knowledge to replace ASME, the NRC decided to modify the former adding environmentally-assisted fatigue multipliers to the traditional usage factors, formally defined as

$$F_\text{en} = \frac{N_\text{air}}{N_\text{water}} \geq 1$$

This way, the environmentally-assisted usage factor for the $j$-th load pair is

$$\text{CUF}_\text{en,j} = U_j \cdot {F_\text{en},j}$$
and the global cumulative usage factor in water is the sum of these partial contributions
This way, the environmentally-assisted usage factor for the $j$-th load pair is $\text{CUF}_\text{en,j} = U_j \cdot {F_\text{en},j}$ and the global cumulative usage factor in water is the sum of these partial contributions

$$\text{CUF}_\text{en} = U_1 \cdot F_{\text{en},1} + U_2 \cdot F_{\text{en},2} + \dots + U_j \cdot F_{\text{en},j} + \dots$${#eq:cufen}

In EPRI’s words:
In EPRI’s words, the general steps for performing an EAF analysis are as follows:

> The general steps for performing an EAF analysis are as follows:
>
> 1. perform an ASME fatigue analysis using fatigue curves for an air
1. perform an ASME fatigue analysis using fatigue curves for an air
environment
> 2. calculate $F_\text{en}$ factors for each transient pair in the fatigue analysis
> 3. apply the $F_\text{en}$ factors to the incremental usage calculated for each
2. calculate $F_\text{en}$ factors for each transient pair in the fatigue analysis
3. apply the $F_\text{en}$ factors to the incremental usage calculated for each
transient pair ($U_j$), to determine the $\text{CUF}_\text{en}$, using\ [@eq:cufen]

Again, if $\text{CUF}_\text{en} < 1$, then the system under study can withstand the assumed cyclic loads. Note that as\ $F_{\text{en},j}$, we can have $\text{CUF} < 1$ and $\text{CUF}_\text{en} > 1$ at the same time.

\medskip

Again, if $\text{CUF}_\text{en} < 1$, then the system under study can withstand the assumed cyclic loads. Note that as\ $F_{\text{en},j}$, we can have $\text{CUF} < 1$ and $\text{CUF}_\text{en} > 1$ at the same time.
The NRC has performed a comprehensive set of theoretical and experimental tests to study and analyse the nature and dependence of the non-dimensional correction factors\ $F_\text{en}$. They found that, for a given material, they depend on:

a. the concentration\ $O(t)$ of dissolved oxygen in the water,
@@ -1275,7 +1079,7 @@ The NRC has performed a comprehensive set of theoretical and experimental tests
c. the strain rate\ $\dot{\epsilon}(t)$, and
d. the content of sulphur\ $S(t)$ in the pipes (only for carbon or low-allow steels).

Apparently it makes no difference whether the environment is composed of either light or heavy water. There are somewhat different sets of non-dimensional analytical expressions that estimate the value of\ $F_{\text{en}}(t)$ as a function of\ $O(t)$, $T(t)$, $\dot{\epsilon}(t)$ and $S(t)$, both in the few revisions of NUREG/CR-6909 and in EPRI’s report \#1025823. Although they are not important now, the actual expressions should be defined and agreed with the plant owner and the regulator. The main result to take into account is that\ $F_{\text{en}}(t)=1$ if\ $\dot{\epsilon}(t)\leq0$, i.e. there are no environmental effects during the time intervals where the material is being compressed.
Apparently it makes no difference whether the environment is composed of either light or heavy water. There are somewhat different sets of non-dimensional analytical expressions that estimate the value of\ $F_{\text{en}}(t)$ as a function of\ $O(t)$, $T(t)$, $\dot{\epsilon}(t)$ and $S(t)$, both in the few revisions of NUREG/CR-6909 and in EPRI’s report\ #1025823. Although they are not important now, the actual expressions should be defined and agreed with the plant owner and the regulator. The main result to take into account is that\ $F_{\text{en}}(t)=1$ if\ $\dot{\epsilon}(t)\leq0$, i.e. there are no environmental effects during the time intervals where the material is being compressed.

Once we have the instantaneous factor\ $F_{\text{en}}(t)$, we need to obtain an average value\ $F_{\text{en},j}$ which should be applied to the\ $j$-th load pair. Again, there are a few different ways of lumping the time-dependent\ $F_{\text{en}}(t)$ into a single $F_{\text{en},j}$ for each interval. Both NRC and EPRI give simple equations that depend on a particular time discretisation of the stress histories that, in my view, are all ill-defined. My guess is that they underestimated they audience and feared readers would not understand the slightly-more complex mathematics needed to correctly define the problem. The result is that they introduced a lot of ambiguities (and even technical errors) just not to offend the maths illiterate. A decision I do not share, and a another reason to keep on learning and practising math.

@@ -1289,41 +1093,39 @@ When faced for the first time with the case study, I have come up with a weighti
Tables of individual environmental correction and usage factors for the NRC/EPRI “EAF Sample Problem 2-Rev.\ 2 (10/21/2011).” The reference method assigns the same\ $F_\text{en}$ to the first two rows whilst the proposed lumping scheme does show a difference
:::::

# Conclusions
## Conclusions

Back in college, we all learned how to solve engineering problems. And once we graduated, we felt we could solve and fix the world (if you did not graduate yet, you will feel it shortly). But there is a real gap between the equations written in chalk on a blackboard (now probably in the form of beamer slide presentations) and actual real-life engineering problems. This chapter introduces a real case from the nuclear industry and starts by idealising the structure such that it has a known analytical solution that can be found in textbooks. Additional realism is added in stages allowing the engineer to develop an understanding of the more complex physics and a faith in the veracity of the finite-element results where theoretical solutions are not available. Even more, a brief insight into the world of evaluation of stress-life fatigue using such results further illustrates the complexities of real-life engineering analysis.

A list of the tips that arose throughout the text:
Back in college, we all learned how to solve engineering problems. And once we graduated, we felt we could solve and fix the world (if you did not graduate yet, you will feel it shortly). But there is a real gap between the equations written in chalk on a blackboard (now probably in the form of beamer slide presentations) and actual real-life engineering problems. This chapter introduces a real case from the nuclear industry and starts by idealising the structure such that it has a known analytical solution that can be found in textbooks. Additional realism is added in stages allowing the engineer to develop an understanding of the more complex physics and a faith in the veracity of the finite-element results where theoretical solutions are not available. Even more, a brief insight into the world of evaluation of stress-life fatigue using such results further illustrates the complexities of real-life engineering analysis. Here is a list of the tips that arose throughout the text:

* use and exercise your imagination
* practise math
* start with simple cases first
* grasp the dependence of results with independent variables
dnl * grasp the dependence of results with independent variables
* keep in mind there are other methods beside finite elements
* take into account that even within the finite element method, there is a wide variety of complexity in the problems that can be solved
* follow the “five whys rule” before compute anything, probably you do not need to
* use engineering judgment and make sure understand the [“wronger than wrong”](https://en.wikipedia.org/wiki/Wronger_than_wrong) concept
dnl * follow the “five whys rule” before compute anything, probably you do not need to
* use engineering judgment and make sure understand Asimov’s [“wronger than wrong”](https://en.wikipedia.org/wiki/Wronger_than_wrong) concept
* play with your favourite FEM solver (mine is [CAEplex](https://caeplex.com)) solving simple cases, trying to predict the results and picturing the stress tensor and its eigenvectors in your imagination
* try to show that even though the principal stresses are not linear with respect to summation, they are linear with respect to multiplication (with pencil and paper)
* prove (with pencil and paper) that even though the principal stresses are not linear with respect to summation, they are linear with respect to multiplication
* grab any stress distribution from any of your FEM projects, compute the iso-stress curves and the draw normal lines to them to get acquainted with SCLs
* first search online for “stress linearisation” (or “linearization” if you want) and then get a copy of ASME\ VIII Div\ 2 Annex 5-A
dnl * first search online for “stress linearisation” (or “linearization” if you want) and then get a copy of ASME\ VIII Div\ 2 Annex 5-A
* keep in mind that FEM solutions lead only to nodal equilibrium but not pointwise equilibrium
* measure the time needed to generate grids of different sizes and kinds with your favourite mesher
* learn this by heart: the complexity of a FEM problem is given mainly by the number of _nodes_, not by the number of elements
* remember that welded materials with different thermal expansion coefficients may lead to fatigue under cyclic temperature changes
* if you have time, try to get out of your comfort zone and do more than what others expect from you (like parametric computations)
* clone the [parametric tee repository](https://bitbucket.org/seamplex/tee), understand how the figures from\ [@sec:parametric] were built and expand them to cover “we might go on...” bullets
* try to find an explanation of the results obtained, just like we did when we explained the parametric curves from\ [@fig:tee-MB ] with two opposing effects which were equal in magnitude around $d_b=5$\ inches
dnl * if you have time, try to get out of your comfort zone and do more than what others expect from you (like parametric computations)
dnl * clone the [parametric tee repository](https://bitbucket.org/seamplex/tee), understand how the figures from\ [@sec:parametric] were built and expand them to cover “we might go on...” bullets
dnl * try to find an explanation of the results obtained, just like we did when we explained the parametric curves from\ [@fig:tee-MB ] with two opposing effects which were equal in magnitude around $d_b=5$\ inches
* think thermal-mechanical plus earthquakes as “bake, break and shake” problems
* understand why the elastic problem of the case study is still linear after all
* keep in mind that finite-elements are a mean to get an engineering solution, not and end by themselves
* learn to write scripts to post-process FEM results (from a script-friendly FEM program)
* work under a [distributed version control](https://en.wikipedia.org/wiki/Distributed_version_control) system such as [Git](https://en.wikipedia.org/wiki/Git), even when just editing input files or writing reports
* clone the [environmental fatigue sample problem repository](https://bitbucket.org/seamplex/cufen) and obtain a nicely-formatted table with the results of the “EAF Sample Problem 2-Rev.\ 2 (10/21/2011)” from\ [@sec:in-air;@sec:in-water].
dnl * clone the [environmental fatigue sample problem repository](https://bitbucket.org/seamplex/cufen) and obtain a nicely-formatted table with the results of the “EAF Sample Problem 2-Rev.\ 2 (10/21/2011)” from\ [@sec:in-air;@sec:in-water].
* do not write ambiguous reports by replacing appropriate mathematical formulae with words just not to offend the illiterate
* try to avoid [proprietary](https://en.wikipedia.org/wiki/Proprietary_software) programs and favour [free and open source](https://en.wikipedia.org/wiki/Free_and_open-source_software) ones.

You can ask for help in our mailing list at <wasora@seamplex.com>. There is a community of engineers willing to help you in case you get in trouble with the repositories, the script or the input files.
dnl You can ask for help in our mailing list at <wasora@seamplex.com>. There is a community of engineers willing to help you in case you get in trouble with the repositories, the script or the input files.

About your favourite FEM program, ask yourself these two questions:

@@ -1332,24 +1134,7 @@ About your favourite FEM program, ask yourself these two questions:
3. Do you trust your favourite FEM program?


And finally, make sure that at the end of the journey from college theory to an actual engineering problem your conscience, is clear knowing that there exists a report with your signature on it. That is why we went to college in the first place.

divert(-1)
# Referenced works

## Books

## Articles

## Reports

## Blog posts
And finally, make sure that at the end of the journey from college theory to an actual engineering problem your conscience, is clear knowing that there exists a report with your signature on it. That is why we all went to college in the first place.

## Websites

esyscmd([[lynx -dump -listonly nafems4.html | grep http | awk '{print $2}' | sort -u | uniq | grep -v wikipedia | awk '{printf(" * <%s>\n", $1)}']])

## Wikipedia

esyscmd([[lynx -dump -listonly nafems4.html | grep http | awk '{print $2}' | sort -u | uniq | grep wikipedia | awk '{printf(" * <%s>\n", $1)}']])
divert(0)
# Concluding Remarks

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- 0
problem.ppl Vedi File

@@ -0,0 +1,18 @@
# problem parameters for
# 12"-inch schedule 100
b = 323.8/2 # external radius [ mm ]
a = b-21.5 # internal radius [ mm ]
l = 2*(b-a) # axial length [ mm ]
n = 4 # number of elements through thickness

E = 200e3 # Young modulus [ MPa ]
nu = 0.3 # Poisson's ratio [ non-dimensional ]
p = 10 # internal pressure [ MPa ]


# definition of analytical solutions for comparison from
# <http://eprints.whiterose.ac.uk/110536/1/art%253A10.1007%252Fs00707-016-1762-7.pdf>
ur(x,y,z) = (p*a**2*sqrt(y**2+z**2))/(E*(b**2-a**2)) * ((1-2*nu)*(1+nu) + (1+nu)*b**2/(y**2+z**2))
sigmal(x,y,z) = 2*nu*p*a**2/(b**2-a**2)
sigmar(x,y,z) = p*a**2/(b**2-a**2) * (1 - b**2/(y**2+z**2))
sigmatheta(x,y,z) = p*a**2/(b**2-a**2) * (1 + b**2/(y**2+z**2))

BIN
real-gen.png Vedi File

Before After
Width: 1200  |  Height: 775  |  Size: 174KB Width: 1436  |  Height: 935  |  Size: 207KB

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upload.sh Vedi File

@@ -1 +1 @@
rsync -avz --progress --exclude=".git" --exclude="refs" . seamplex.com:html/jimbo/docs/nafems4
rsync -avz --progress --exclude=".git" --exclude="refs" . seamplex.com:html/milhouse/html/docs/nafems4

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