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@@ -243,7 +243,6 @@ Finally, when then uncertainties associated to the parameters, methods and model |
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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 swifting the input parameters so as to obtain a distribution of possible outcomes. |
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**------** |
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## Five whys {#sec:five} |
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@@ -490,12 +489,15 @@ Let us both (i.e. you and me) make an experiment. Grab a FEM program of your cho |
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Now we are going to create and compare three load cases: |
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A. Pure normal loads (<https://caeplex.com/p?d8f>) |
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B. Pure shear loads (<https://caeplex.com/p?b494>) |
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C. The combination of A & B (<https://caeplex.com/p?989>) |
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a. Pure normal loads (<https://caeplex.com/p?d8f>) |
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b. Pure shear loads (<https://caeplex.com/p?b494>) |
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c. The combination of A & B (<https://caeplex.com/p?989>) |
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The loads in each cases are applied to the three remaining faces, namely “right”, “back” and “top,” and their magnitude in Newtons are: |
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````{=latex} |
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\begin{center} |
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\rowcolors{2}{black!10}{black!0} |
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\begin{tabular}{l|ccc|ccc|ccc} |
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& |
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\multicolumn{3}{c|}{face “right” ($x>0$)} & |
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@@ -519,60 +521,134 @@ Case A, pure normal & +10 & 0 & 0 & 0 & +20 & 0 & 0 & 0 & +30 \\ |
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Case B, pure shear & 0 & +15 & -15 & +25 & 0 & -5 & -15 & +25 & +30 \\ |
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Case C, combination & +10 & +15 & -15 & +25 & +20 & -5 & -15 & +25 & +30 \\ |
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\end{tabular} |
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\end{center} |
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```` |
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````{=html} |
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<table> |
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<tr> |
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<th></th> |
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<th colspan="3">face “right” ($x>0$)</th> |
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<th colspan="3">face “right” ($x>0$)</th> |
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<th colspan="3">face “right” ($x>0$)</th> |
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</tr> |
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<tr> |
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<th></th> |
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<th>$F_x$</th> |
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<th>$F_y$</th> |
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<th>$F_z$</th> |
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<th>$F_x$</th> |
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<th>$F_y$</th> |
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<th>$F_z$</th> |
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<th>$F_x$</th> |
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<th>$F_y$</th> |
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<th>$F_z$</th> |
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</tr> |
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<tr> |
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<td>Case A, pure normal</td> |
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<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> |
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</tr> |
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<tr> |
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<td>Case B, pure shear</td> |
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<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> |
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</tr> |
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<tr> |
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<td>Case C, combination</td> |
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<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> |
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</tr> |
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</table> |
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```` |
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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 1mm², the usual sorting leads to |
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In the first case, the principal stresses are uniform and equal to the three normal loads. As the forces are in Newton and the area of each face of the cube is 1mm^2, the usual sorting leads to |
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divert(-1) |
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$$ |
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\begin{align*} |
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\sigma_{1A} &= 30 \text{MPa} \\ |
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\sigma_{2A} &= 20 \text{MPa} \\ |
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\sigma_{3A} &= 10 \text{MPa} \\ |
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\sigma_{1A} &= 30~\text{MPa} \\ |
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\sigma_{2A} &= 20~\text{MPa} \\ |
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\sigma_{3A} &= 10~\text{MPa} \\ |
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\end{align*} |
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$$ |
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divert(0) |
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In the second case, the principal stresses are not uniform and have a non-trivial distribution. The results obtained in CAeplx |
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$$ |
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\sigma_{1A} = 30~\text{MPa} |
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$$ |
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$$ |
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\sigma_{2A} = 20~\text{MPa} |
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$$ |
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$$ |
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\sigma_{3A} = 10~\text{MPa} |
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$$ |
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::::: {#fig:cube} |
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{#fig:cube-shear width=50%} |
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{#fig:cube-full width=50%} |
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podemos sumar los desplazamientos y los tensores de tensiones, pero ojo que las tensiones principales no son lineales a la suma (si a la multiplicacion) |
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Spatial distribution of principal stress\ 3 for cases\ B and\ C. If linearity applied, case\ C would be equal to case\ B plus an constant. |
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::::: |
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In the second case, the principal stresses are not uniform and have a non-trivial distribution. Indeed, the distribution of\ $\sigma_3$ obtained by CAeplex is shown in\ [@fig:cube-shear]. |
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Now, if we indeed were facing a fully linear problem then the results of the sum of two inputs would be equal to the sum of the individual inputs. And\ [@fig:cube-full], which shows the principal stress\ 3 of case\ C is not the result from case\ B plus any of the three constants from case\ A. Had it been, the color distribution would have been _exactly_ the same as the scale goes automatically from the most negative value in blue to the most positive value in red. And 7+30\ $\neq$ 33. Alas, it seems that there exists some kind of unexpected non-linearity (the feared master-pet interaction?) that prevents us from from fully splitting the problem into simpler chunks. |
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So what is the source of this unexpected non-linear effect in an otherwise nice and friendly linear formulation? Well, probably you already know it because after all it is almost high-school mathematics. But I learned it way after college when facing a real engineering problem and not just back-of-the-envelope pencil-and-paper trivial exercises. |
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Recall that principal stresses are the eigenvalues of the stress tensor. And the fact that in a linear elastic formulation means that the stress tensor of case\ C above is the sum of the individual stress tensors from cases\ A and B does not mean that their eigenvalues can be summed (think about it!). Again, imagine the eigenvalues and eigenvectors of cases A & B. Got it? Good. Now imagine the eigenvalues and eigenvectors for case\ C. Should they sum up? No, they should not! Let us make another experiment, this time with matrices using [Octave](https://www.gnu.org/software/octave/) or whatever matrix-friendly program you want. |
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octave:32> A = rand(3); A = A*A' |
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First, let us create a 3$\times$3 random matrix $R$ and then multiply it by its transpose\ $R^T$ to obtain a symmetric matrix\ $A$ (recall that the stress tensor is symmetric): |
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```octave |
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octave> R = rand(3); A = R*R' |
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A = |
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2.08711 1.40929 1.31108 |
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1.40929 1.32462 0.57570 |
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1.31108 0.57570 1.09657 |
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``` |
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Do the same to obtain another 3$\times$3 symmetric matrix\ B: |
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octave:33> B = rand(3); B = B*B' |
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```octave |
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octave> R = rand(3); B = R*R' |
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B = |
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1.02619 0.73457 0.56903 |
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0.73457 0.53386 0.37772 |
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0.56903 0.37772 0.53141 |
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``` |
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Now compute the sum of the eigenvalues first and then the eigenvalues of the sum: |
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octave:34> eig(A)+eig(B) |
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```octave |
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octave> eig(A)+eig(B) |
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ans = |
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0.0075113 |
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0.8248395 |
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5.7674016 |
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octave:35> eig(A+B) |
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octave> eig(A+B) |
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ans = |
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0.049508 |
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0.782990 |
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5.767255 |
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``` |
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octave:36> |
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divert(0) |
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Did I convince you? More or less, right? The third eigenvalue seems to fit. Let us not throw all of our beloved linearity away and dig in further into the subject. There are still two important issues to discuss which can be easily addressed using fresh-year linear algebra (remember, do not fear math!). First of all, even though principal stresses are not linear with respect to the sum they are linear with respect to pure multiplication. Once more, think what happens to the the eigenvalues and eigenvectors of a single stress tensor as all its elements are scaled up or down by a real scalar. They are the same! So, for example, the [Von Mises stress](https://en.wikipedia.org/wiki/Von_Mises_yield_criterion) (which is a combination of the principal stresses) of a beam loaded with a force\ $\alpha \cdot \vec{F}$ is\ $\alpha$ times the stress of the beam loaded with a force\ $\vec{F}$. Please test this hypothesis by playing with your favorite FEM solver an play. Or even better, take a look at the stress invariants $I_1$, $I_2$ and $I_3$ (you can search online or peek into the source code of [Fino](https://www.seamplex.com/fino) and grep for the routine called `fino_compute_principal_stress()`) and see (using paper and pencil!) how they scale up if the individual elements of the stress tensor are scaled by a real factor\ $\alpha$. |
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I did not know this in college! I learned it the hard way. |
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The other issue is that even though in general the eigenvalues of the sum of two matrices are not the same as the eigenvalues of the matrix sum, there are some cases when they are. In effect, if two matrices\ $A$ and\ $B$ commute, i.e. their product is commutative |
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cantilever beam, principal stresses, linearity of von mises |
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$$ |
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A \cdot B = B \cdot A |
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$$ |
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then the sums of their eigenvalues are equal to the eigenvalues of the sums. In order for this to happen, both\ $A$ and\ $B$ need to be diagonalizable using the same basis. That is to say, the diagonalizing matrix\ $P$ such that $P^{-1} A P$ is diagonal is the same that renders\ $P^{-1} B P$ also diagonal. What does this mecanically mean? Well, if case\ A has loads that are somehow “independent” from the ones in case\ B, then the principal stresses of the combination might be equal to the sum of the individual principal stresses. A notable case is for example a beam that is loaded vertically in case\ A and horizontally in case\ B. I dare you to grab your FEM program one more time, run a test and picture the eigenvalues and eigenvectors of the three cases in your head. |
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# The infinite pipe revisited after college |
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@@ -597,13 +673,16 @@ complete vs incomplete (hexa) |
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### ASME stress linearization |
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### ASME stress linearization (not linearity!) |
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## Two (or more) materials |
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open source! |
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que hacen los programas? NADIE SABE |
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### Young and Poisson |
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two cubes |
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@@ -630,6 +709,7 @@ Back in College, we all learned how to solve engineering problems. But there is |
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* within the finite element method, there is a wide variety of complexity in the problems that can be solved |
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* follow the “five whys rule” before compute anything, probably you do not need to |
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* use engineering judgment and make sure understand the [“wronger than wrong”](https://en.wikipedia.org/wiki/Wronger_than_wrong) concept |
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* play with your favourite FEM solver (mine is <https://caeplex.com>) solving simple cases, trying to predict the results and picturing the stress tensor and its eigenvectors in your imagination |
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dnl errors and uncertainties: model parameters (is E what we think? is the material linear?), geometry (does the CAD represent the reality?) equations (any effect we did not have take account), discretization (how well does the mesh describe the geometry?) |
gtheler
hace 7 años
