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@@ -412,11 +412,11 @@ As a nuclear engineer, I learned (theoretically in college but practically after |
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Therefore, it is worth to take some time to think about what we need to do, what our choices are to build finite-element [models](https://www.seamplex.com/blog/say-modeling-not-simulation.html) and which one is the most convenient in terms of costs and efficiency. First of all, we need to define which transients are going to be taken into account. For the current imaginary case study, we define that the piping system from [@fig:cad-figure] will be subject to the the four simple (and again imaginary) time histories for the internal pressure\ $p$ and the fluid temperature\ $T$ as a function of time shown in [@fig:pt].^[Actual real piping systems might be subject to dozens of more complex transients.] |
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::::: {#fig:pt} |
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{#fig:pt1 width=95%} |
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{#fig:pt1 width=95%} |
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{#fig:pt2 width=95%} |
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{#fig:pt3 width=95%} |
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{#fig:pt3 width=95%} |
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{#fig:pt4 width=95%} |
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@@ -1033,9 +1033,9 @@ We then proceed to “shake” the pipes. That is to say, we obtain a distribute |
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Finally we attempt to “break” the pipes successively solving many steady-state (i.e. quasi-static) elastic problems for different times\ $t$ of each of the operational transients from [@fig:pt]. The principal stresses at the internal points of two SCLs for the four transients juxtaposed into a single time history are shown in [@fig:sigmas]. Some remarks about this step: |
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:::: {#fig:MB-scl} |
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{#fig:MB-scl-1} |
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{#fig:MB-scl-2} |
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Juxtaposition of the linearized MB principal stresses at two SCLs. |
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:::: |
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@@ -1132,91 +1132,101 @@ When\ $\text{CUF} < 1$, the part under analysis can withstand the proposed cycli |
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This cryptic paragraph is a clear example of stuff that cannot be learned at college. No matter how good your university is, there is no way to cover all theories and methodologies which a mechanical engineer could need in his or her professional life. |
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dnl ------------------------------------------------------- |
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Let’s start by taking into account the juxtaposed stress histories from [@fig:mb-scl]. ASME NB-3216 requires to take the $\text{MB}_1-\text{MB}_3$ difference and perform some slight corrections for possible plastic and temperature effects which are out of the scope of this case study. This modified stress history $\text{MB}^{\prime}_{31}$, which is shown in [@fig:extrema-4] for the SCL\ #1 along with the temperature and pressure transients for reference, is used to evaluate fatigue resistance “in air” as follows. |
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{#fig:extrema-1} |
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Todo referido a sigma(0) |
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| $t$ | $\Delta \text{MB}^{\prime}_{31}$ | $\Delta[\sigma_3 - \sigma_1]$ | Cycles | Extrema | |
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|:-----:|-----------------------------------:|--------------------------------:|:-------:|:---------:| |
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| 0 | 0.0 | 0.0 | 200 | initial |
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| 352 | -381.7 | -298.5 | 200 | min |
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| 3131 | -2.1 | -3.2 | 200 | max |
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| 3262 | -301.0 | -270.6 | 100 | min |
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| 4812 | -146.9 | -133.9 | 100 | max |
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| 4823 | -284.0 | -199.5 | 100 | min |
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| 6523 | -330.0 | -284.2 | 100 | min |
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| 6712 | -282.5 | -253.7 | 100 | final |
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: Extrema of the juxtaposed stress history of [@fig:extrema-1] |
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It consists of a typical vessel nozzle with attached piping as illustrated in\ [@fig:axi-inches-3d]. These components are subject to four transients\ $k=1,2,3,4$ that give rise to linearised stress histories (slightly modified according to NB-3216.2) which are given as individual stress values juxtaposed so as to span a time range of about 36,000 seconds ([@fig:nureg1]). As the time steps is not constant, each stress value has an integer index\ $i$ that uniquely identifies it: |
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| $k$ | Time range [s] | Index range | Cycles\ $n_k$ | |
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|:-----:|:--------------:|:-----------:|:----------------:| |
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| 1 | 0--3210 | 1--523 | 20 | |
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| 2 | 3210--6450 | 524--959 | 50 | |
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| 3 | 6450--9970 | 960--1595 | 20 | |
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| 4 | 9970--35971 | 1596--2215 | 100 | |
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Let us start by taking into account the juxtaposed stress histories from [@fig:MC-scl-1]. ASME NB-3216 requires to take the $\text{MB}_1-\text{MB}_3$ difference with respect to the initial stress so as to start with a zero value. This differential stress history $\Delta \text{MB}^{\prime}_{31}$, which is shown in [@fig:extrema-1] for the SCL\ #1 along with the temperature and pressure transients for reference, is used to evaluate fatigue resistance “in air” as follows. |
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divert(-1) |
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| $t$ | $\Delta \text{MB}^{\prime}_{31}$ | $\Delta[\sigma_3 - \sigma_1]$ | Transient | Cycles | Extrema | |
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|:-----:|-----------------------------------:|--------------------------------:|:---------:|:--------:|:---------:| |
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| 0 | 0.0 | 0.0 | #1 | 250 | initial | |
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| 352 | -381.7 | -298.5 | #1 | 250 | min | |
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| 3131 | -2.1 | -3.2 | #2 | 200 | max | |
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| 3262 | -301.0 | -270.6 | #3 | 100 | min | |
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| 4812 | -146.9 | -133.9 | #3 | 100 | max | |
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| 4823 | -284.0 | -199.5 | #4 | 100 | min | |
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| 6523 | -330.0 | -284.2 | #4 | 100 | min | |
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| 6712 | -282.5 | -253.7 | #4 | 100 | final | |
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: Extrema of the juxtaposed stress history of [@fig:extrema-1] {#tbl:extrema} |
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divert(0) |
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{#fig:extrema-1} |
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```{=latex} |
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\rowcolors{1}{black!0}{black!10} |
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\begin{table} |
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\begin{center} |
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\begin{tabular}{ |
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c |
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S[table-format=3.1] S[table-format=3.1] |
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c |
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S[table-format=3.0] |
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c |
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} |
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\toprule |
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{$t$} & |
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{$\Delta \text{MB}^{\prime}_{31}$} & {$\Delta[\sigma_3 - \sigma_1]$ } & |
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{Transient} & |
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{Cycles} & |
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{Extrema} \\ |
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\midrule |
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0 & 0.0 & 0.0 & \#1 & 250 & initial \\ |
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352 & -381.7 & -298.5 & \#1 & 250 & min \\ |
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3131 & -2.1 & -3.2 & \#2 & 200 & max \\ |
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3262 & -301.0 & -270.6 & \#3 & 100 & min \\ |
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4812 & -146.9 & -133.9 & \#3 & 100 & max \\ |
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4823 & -284.0 & -199.5 & \#4 & 100 & min \\ |
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6523 & -330.0 & -284.2 & \#4 & 100 & min \\ |
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6712 & -282.5 & -253.7 & \#4 & 100 & final \\ |
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\bottomrule |
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\end{tabular} |
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\end{center} |
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\caption{\label{tbl:extrema} Extrema of the juxtaposed stress history of [@fig:extrema-1]} |
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\end{table} |
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``` |
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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. |
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To compute the global usage factor, we first need to find all the combinations of local extrema pairs and then sort them in decreasing order of stress difference. For example, the largest stress amplitude is found between $i=447$ and $i=694$. The second one is 447--699. Then 699--1020, 699--899, etc. For each of these pairs, defined by the indexes\ $i_{1,j}$ and $i_{2,j}$, a partial usage factor\ $U_j$ should computed. The stress amplitude\ $S_{\text{alt},j}$ which should be used to enter into the $S$-$N$ curve is |
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First all the local extrema (i.e. whether a minimum or a maximum) need to be identified. [@Tbl:extrema] shows the times at which these occur, the stresses associated to them and the number of cycles that each of them is expected to occur. The initial and final stresses are also taken into account. It also illustrates the difference between the linearised stress and the plain Tresca scalar stress. To compute the global usage factor, we need to find all the combinations of these local extrema pairs and then sort them in decreasing order of stress difference. For example, the largest stress amplitude is found between $t=0$ and $t=352$ (this last instant contains the seismic load!). The second one is 352--3131. Then 0--6523, 3131--6523, etc. For each of these pairs, defined by the times\ $t_{1,j}$ and $t_{2,j}$, a partial usage factor\ $U_j$ should computed. The stress amplitude\ $S_{\text{alt},j}$ which should be used to enter into the $S$-$N$ curve is |
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$$ |
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S_{\text{alt},j} = \frac{1}{2} \cdot k_{e,j} \cdot \left| MB^\prime_{i_{1,j}} - MB^\prime_{i_{2,j}} \right| \cdot \frac{E_\text{SN}}{E(T_{\text{max}_j})} |
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S_{\text{alt},j} = \frac{1}{2} \cdot k_{\nu,j} \cdot k_{e,j} \cdot \left| \Delta MB^\prime_{t_{1,j}} - \Delta MB^\prime_{t_{2,j}} \right| \cdot \frac{E_\text{SN}}{E(T_{\text{max}_j})} |
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$$ |
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\noindent where $k_e$ is a plastic correction factor for large loads (NB-3228.5), $E_\text{SN}$ is the Young’s Modulus used to create the $S$-$N$ curve (provided in the ASME fatigue curves) and\ $E(T_{\text{max}_j})$ is the material’s Young’s Modulus at the maximum temperature within the\ $j$-th interval. |
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\noindent where $k_\nu$ and $k_e$ are plastic correction factor for large loads (part VIII div 2 sec 5.5.3.2 and part III NB-3228.5), $E_\text{SN}$ is the Young’s Modulus used to create the $S$-$N$ curve (provided in the ASME fatigue curves) and\ $E(T_{\text{max}_j})$ is the material’s Young’s Modulus at the maximum temperature within the\ $j$-th interval. |
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We now need to comply with ASME’s obscure note about the number of cycles to assign a proper value of\ $n_j$. Back to the largest pair 447-694, we see that 447 belongs to transient\ #1 which has assigned 20 cycles and 694 belongs to the earthquake with 5 cycles. Therefore $n_1=5$, and the cycles associated to each index are decreased in five. So $i=694$ disappears and the number of cycles associated to $i=447$ are decreased from 20 to 15. The second largest pair is now 447-699, with 15 (because we just spent 5 in the first pair) and 50 cycles respectively. Now $n_2=15$, point 447 disappears and 699 remains with 35 cycles. The next pair is 699-1020, with number of cycles 35 and 20 so $n_3=20$, point 1020 disappears and 699 remains with 15 cycles. And so on, down to the last pair. |
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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 automate 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. |
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We are now in a position where we can comply with ASME’s obscure note about the number of cycles to assign a proper value of\ $n_j$. Starting with the largest pair 0--352, we see that both extrema belong to transient #1 which has 250 cycles. This one is easy, because we associate directly $n_1=250$ and both of these times “dissappear” as they already consumed all their cycles. The second largest pair was 352--3131 but 352 has just vanished so it is not considered anymore. The following is 0--6523 but zero also consumed all its cycles, so this pair is also discarded. The next is now 331--6523 where the first one belongs to transient #2 (200 cycles) and the latter to transient #4 (100 cycles). We assign $n_2=\min(200,100)=100$ and subtract 100 to both the cycles remaining to each time. Point 6523 dissapears as it consumed all its initial cycles and 3131 remains with 100 cycles. The next pair is 3131--3262, with number of cycles 100 (because we just took away 100 out of the initial 200) and 150 so $n_3=100$, point 3131 disappears and 3262 remains with 50 cycles. And so on, down to the last pair. [@Tbl:table-cuf] shows the results of applying this algorithm to all the extrema in SCL\ #1. The columns try to match the “official” solution of the US\ NRC [@nrc] to a sample problem proposed by the Electric Power Research Institute [@epri], which is shown in [@tbl:cuf-nrc]. |
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::::: {#fig:nureg} |
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{#fig:nureg1 width=70%} |
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dnl 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. |
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{#fig:nureg2 width=70%} |
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dnl ::::: {#tbl:cuf} |
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{#tbl:table-cuf width=100%} |
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{#fig:nureg3 width=70%} |
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![Results by reported [@nrc] to the sample problem proposed in [@epri]](cuf-nrc.png){#tbl:cuf-nrc width=100%} |
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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) |
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::::: |
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::::: {#fig:cuf} |
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{#fig:cuf-nrc width=100%} |
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dnl Tables of individual usage factors |
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dnl ::::: |
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{#fig:cuf-seamplex width=100%} |
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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. |
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::::: |
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Why all these details? Not because I want to teach you how to perform fatigue evaluations just reading this section without fully understanding the ASME code, taking college courses on material fatigue, reading books on the subject and even asking 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 automate 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. See [@sec:online] to obtain the set of scripts that detected, matched and sorted the extrema and built [@fig:extrema-1] and [@tbl:extrema] automatically. |
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### In water (NRC’s extension) {#sec:in-water} |
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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 |
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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 by adding environmentally-assisted fatigue multipliers to the traditional usage factors, formally defined as |
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$$F_\text{en} = \frac{N_\text{air}}{N_\text{water}} \geq 1$$ |
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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 |
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Thus, the environmentally-assisted usage factor for the $j$-th load pair is |
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$$\text{CUF}_\text{en,j} = U_j \cdot F_{\text{en},j}$$ |
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\noindent and the global cumulative usage factor in water is the sum of these partial contributions |
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$$\text{CUF}_\text{en} = U_1 \cdot F_{\text{en},1} + U_2 \cdot F_{\text{en},2} + \dots + U_j \cdot F_{\text{en},j} + \dots$${#eq:cufen} |
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In EPRI’s words, the general steps for performing an EAF analysis are as follows: |
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In EPRI’s words, the general steps for performing an environmentally-assisted fatigue (EAF) analysis are as follows [@epri]: |
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1. perform an ASME fatigue analysis using fatigue curves for an air |
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environment |
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@@ -1224,16 +1234,21 @@ environment |
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3. apply the $F_\text{en}$ factors to the incremental usage calculated for each |
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transient pair ($U_j$), to determine the $\text{CUF}_\text{en}$, using\ [@eq:cufen] |
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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. |
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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: |
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Again, if $\text{CUF}_\text{en} < 1$, then the system under study can withstand the assumed cyclic loads. Note that as\ $F_{\text{en},j}>1$, it might be possible to have $\text{CUF} < 1$ and $\text{CUF}_\text{en} > 1$ at the same time. |
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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\ $F_\text{en}$ [@nrc]. They found that, for a given material, they depend on: |
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a. the concentration\ $O(t)$ of dissolved oxygen in the water, |
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b. the temperature\ $T(t)$ of the pipe, |
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c. the strain rate\ $\dot{\epsilon}(t)$, and |
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d. the content of sulphur\ $S(t)$ in the pipes (only for carbon or low-allow steels). |
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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. |
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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 $S(t)$. 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. |
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Without further diving into another level of mathematical complexities and raising a plethora of detailed technical considerations, it is enough to directly show what the results of this EAF analysis for the imaginary test case in [@tbl:table-cufen]. Actually, SCL \#1 was chosen throughout this section because the min/max extrema was simple and thus the explanation of the procedure was easier. It is SCL \#4 the one that has a larger cumulative usage factor |
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{#tbl:table-cufen width=45%} |
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divert(-1) |
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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 their 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. |
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When faced for the first time with the case study, I have come up with a weighting method that I claim is less ill-defined (it still is for border-line cases) and which the plant owner accepted as valid. [@Fig:cufen] shows the reference results of the problem (based on computing two correction factors and then taking the maximum) and the ones obtained with the proposed method (by computing a weighted integral between the valley and the peak). Note how in\ [@fig:cufen-nrc], pairs 694-447 and 699-447 have the same\ $F_\text{en}$ even though they are (marginally) different. The results from\ [@fig:cufen-seamplex] give two (marginally) different correction factors. |
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@@ -1245,6 +1260,8 @@ When faced for the first time with the case study, I have come up with a weighti |
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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 |
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::::: |
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divert(0) |
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## Conclusions |
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gtheler
5 years ago
