added myst markdown too

book/week_13/session_1/intro.ipynb

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  "cells": [
   {
    "cell_type": "markdown",
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-   "metadata": {
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+   "metadata": {},
    "source": [
     "```{index} Transformations; Class exercise using analytical formulas\n",
     "```\n",
@@ -14,6 +12,10 @@
     ":text: \".ipynb\"\n",
     "```\n",
     "\n",
+    "```{custom_download_link} intro.md\n",
+    ":text: \".md\"\n",
+    "```\n",
+    "\n",
     "(lesson13.1)=\n",
     "# Lesson November 25th\n",
     "\n",
@@ -78,7 +80,7 @@
   {
    "cell_type": "code",
    "execution_count": null,
-   "id": "6c66e8e3",
+   "id": "41f3415c",
    "metadata": {
     "tags": [
      "remove-cell"
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  ],
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-   "formats": "ipynb,py:percent"
+   "cell_markers": "\"\"\""
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    "display_name": "base",

book/week_13/session_1/intro.md

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+---
+jupytext:
+  cell_markers: '"""'
+  text_representation:
+    extension: .md
+    format_name: myst
+    format_version: 0.13
+    jupytext_version: 1.16.1
+kernelspec:
+  display_name: base
+  language: python
+  name: python3
+---
+
+```{index} Transformations; Class exercise using analytical formulas
+```
+
+```{custom_download_link} intro.ipynb
+:text: ".ipynb"
+```
+
+```{custom_download_link} intro.md
+:text: ".md"
+```
+
+(lesson13.1)=
+# Lesson November 25th
+
+During today's lesson you'll work on a complex exercise on the topic of the Transforming tensors. Please ask your questions regarding the [homework](homework13.1) as well!
+
+## Exercise Transforming tensors
+
+Given is the following structure and cross-section:
+
+```{figure} intro_data/structure.svg
+:align: center
+```
+
+1. Find the relevant cross-sectional properties.
+2. Find the normal and shear stresses just below $\text{G}$, in $\text{H}$, in $\text{I}$ and just right of $\text{C}$ in cross-section $\text{A}$.
+3. Find the principal values of the stresses in the points just below $\text{G}$, in $\text{H}$, in $\text{I}$ and just right of $\text{C}$ in cross-section $\text{A}$.
+
+````{admonition} Solution assignment 1
+:class: tip, dropdown
+
+Normal force centre is given by:
+```{figure} intro_data/NC.svg
+:align: center
+```
+
+- $A \approx  17500 \text{ mm}^2$
+- $I_{zz} \approx 655 \cdot 10^6 \text{ mm}^4$
+
+````
+
+````{admonition} Solution assignment 2
+:class: tip, dropdown
+
+- $\sigma_\text{just below G} = +6.73 \text{ MPa}$
+- $\tau_\text{just below G} = +0.164 \text{ MPa}$
+- $\sigma_\text{H} = +6.73 \text{ MPa}$
+- $\tau_\text{H} = 0 \text{ MPa}$
+- $\sigma_\text{I} = -2 \text{ MPa}$
+- $\tau_\text{I} = +0.35 \text{ MPa}$
+- $\sigma_\text{just right of C} = -8.53 \text{ MPa}$
+- $\tau_\text{just right of C} = -0.12 \text{ MPa}$
+
+````
+
+````{admonition} Solution assignment 3
+:class: tip, dropdown
+
+- $\sigma_\text{1, just below G} = +6.73 \text{ MPa}$
+- $\sigma_\text{2, just below G} = -0.0040 \text{ MPa}$
+- $\sigma_\text{1, H} = +6.73 \text{ MPa}$
+- $\sigma_\text{2, H} = 0 \text{ MPa}$
+- $\sigma_\text{1, I} = 0.59 \text{ MPa}$
+- $\sigma_\text{2, I} = -2.06 \text{ MPa}$
+- $\sigma_\text{1, just right of C} = 0.0018 \text{ MPa}$
+- $\sigma_\text{2, just right of C} = -8.5 \text{ MPa}$
+
+````
+
+Test of adding more text
+
+```{code-cell}
+:tags: [remove-cell]
+
+Izz = 655e6
+
+tau_G = -3e3 * -250*10*286 / 20 / Izz
+print(tau_G)
+
+tau_C = -3e3 * -250*20*(286-500) / 40 / Izz
+print(tau_C)
+
+sigma_G = +6.73
+tau_G = -tau_G
+sigma_G_1 = 1/2 * (sigma_G + 0) + ((1/2 * (sigma_G + 0))**2 + tau_G**2)**0.5
+sigma_G_2 = 1/2 * (sigma_G + 0) - ((1/2 * (sigma_G + 0))**2 + tau_G**2)**0.5
+print(sigma_G_1, sigma_G_2)
+
+sigma_H = +6.73
+tau_G = 0
+sigma_H_1 = 1/2 * (sigma_H + 0) + ((1/2 * (sigma_H + 0))**2 + tau_G**2)**0.5
+sigma_H_2 = 1/2 * (sigma_H + 0) - ((1/2 * (sigma_H + 0))**2 + tau_G**2)**0.5
+print(sigma_H_1, sigma_H_2)
+
+sigma_I = -2
+tau_I = +0.35
+sigma_I_1 = 1/2 * (sigma_I + 0) + ((1/2 * (sigma_I + 0))**2 + tau_I**2)**0.5
+sigma_I_2 = 1/2 * (sigma_I + 0) - ((1/2 * (sigma_I + 0))**2 + tau_I**2)**0.5
+print(sigma_I_1, sigma_I_2)
+
+sigma_C = -8.53
+tau_C = -tau_C
+sigma_C_1 = 1/2 * (sigma_C + 0) + ((1/2 * (sigma_C))**2 + tau_C**2)**0.5
+sigma_C_2 = 1/2 * (sigma_C + 0) - ((1/2 * (sigma_C))**2 + tau_C**2)**0.5
+print(sigma_C_1, sigma_C_2)
+```

book/week_13/session_1/intro.py

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 # jupyter:
 #   jupytext:
 #     cell_markers: '"""'
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 #     text_representation:
 #       extension: .py
 #       format_name: percent
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 :text: ".ipynb"
 ```
 
+```{custom_download_link} intro.md
+:text: ".md"
+```
+
 (lesson13.1)=
 # Lesson November 25th