diff --git a/book/_config.yml b/book/_config.yml index 53977b9..b42816d 100644 --- a/book/_config.yml +++ b/book/_config.yml @@ -39,6 +39,10 @@ sphinx: chtml: { mtextInheritFont: true # To typeset text within math prettier } + nb_custom_formats: + .py: + - jupytext.reads + - fmt: py local_extensions: apastyle: _ext/ bracket_citation_style: _ext/ diff --git a/book/week_13/session_1/intro.ipynb b/book/week_13/session_1/intro.ipynb deleted file mode 100644 index 690b583..0000000 --- a/book/week_13/session_1/intro.ipynb +++ /dev/null @@ -1,154 +0,0 @@ -{ - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "```{index} Transformations; Class exercise using analytical formulas\n", - "```\n", - "\n", - "(lesson13.1)=\n", - "# Lesson November 25th\n", - "\n", - "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!\n", - "\n", - "## Exercise Transforming tensors\n", - "\n", - "Given is the following structure and cross-section:\n", - "\n", - "```{figure} intro_data/structure.svg\n", - ":align: center\n", - "```\n", - "\n", - "1. Find the relevant cross-sectional properties.\n", - "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 $\\text{A}$.\n", - "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 $\\text{A}$.\n", - "\n", - "````{admonition} Solution assignment 1\n", - ":class: tip, dropdown\n", - "\n", - "Normal force centre is given by:\n", - "```{figure} intro_data/NC.svg\n", - ":align: center\n", - "```\n", - "\n", - "- $A \\approx 17500 \\text{ mm}^2$\n", - "- $I_{zz} \\approx 655 \\cdot 10^6 \\text{ mm}^4$\n", - "\n", - "````\n", - "\n", - "````{admonition} Solution assignment 2\n", - ":class: tip, dropdown\n", - "\n", - "- $\\sigma_\\text{just below G} = +6.73 \\text{ MPa}$\n", - "- $\\tau_\\text{just below G} = +0.164 \\text{ MPa}$\n", - "- $\\sigma_\\text{H} = +6.73 \\text{ MPa}$\n", - "- $\\tau_\\text{H} = 0 \\text{ MPa}$\n", - "- $\\sigma_\\text{I} = -2 \\text{ MPa}$\n", - "- $\\tau_\\text{I} = +0.35 \\text{ MPa}$\n", - "- $\\sigma_\\text{just right of C} = -8.53 \\text{ MPa}$\n", - "- $\\tau_\\text{just right of C} = -0.12 \\text{ MPa}$\n", - "\n", - "````\n", - "\n", - "````{admonition} Solution assignment 3\n", - ":class: tip, dropdown\n", - "\n", - "- $\\sigma_\\text{1, just below G} = +6.73 \\text{ MPa}$\n", - "- $\\sigma_\\text{2, just below G} = -0.0040 \\text{ MPa}$\n", - "- $\\sigma_\\text{1, H} = +6.73 \\text{ MPa}$\n", - "- $\\sigma_\\text{2, H} = 0 \\text{ MPa}$\n", - "- $\\sigma_\\text{1, I} = 0.59 \\text{ MPa}$\n", - "- $\\sigma_\\text{2, I} = -2.06 \\text{ MPa}$\n", - "- $\\sigma_\\text{1, just right of C} = 0.0018 \\text{ MPa}$\n", - "- $\\sigma_\\text{2, just right of C} = -8.5 \\text{ MPa}$\n", - "\n", - "````" - ] - }, - { - "cell_type": "code", - "execution_count": 1, - "metadata": { - "tags": [ - "remove-cell" - ] - }, - "outputs": [ - { - "name": "stdout", - "output_type": "stream", - "text": [ - "0.16374045801526718\n", - "-0.12251908396946565\n", - "6.733981439187675 -0.0039814391876737965\n", - "6.73 0.0\n", - "0.059481005020854516 -2.0594810050208547\n", - "0.0017594173959141202 -8.531759417395914\n" - ] - } - ], - "source": [ - "Izz = 655e6\n", - "\n", - "tau_G = -3e3 * -250*10*286 / 20 / Izz\n", - "print(tau_G)\n", - "\n", - "tau_C = -3e3 * -250*20*(286-500) / 40 / Izz\n", - "print(tau_C)\n", - "\n", - "sigma_G = +6.73\n", - "tau_G = -tau_G\n", - "sigma_G_1 = 1/2 * (sigma_G + 0) + ((1/2 * (sigma_G + 0))**2 + tau_G**2)**0.5\n", - "sigma_G_2 = 1/2 * (sigma_G + 0) - ((1/2 * (sigma_G + 0))**2 + tau_G**2)**0.5\n", - "print(sigma_G_1, sigma_G_2)\n", - "\n", - "sigma_H = +6.73\n", - "tau_G = 0\n", - "sigma_H_1 = 1/2 * (sigma_H + 0) + ((1/2 * (sigma_H + 0))**2 + tau_G**2)**0.5\n", - "sigma_H_2 = 1/2 * (sigma_H + 0) - ((1/2 * (sigma_H + 0))**2 + tau_G**2)**0.5\n", - "print(sigma_H_1, sigma_H_2)\n", - "\n", - "sigma_I = -2\n", - "tau_I = +0.35\n", - "sigma_I_1 = 1/2 * (sigma_I + 0) + ((1/2 * (sigma_I + 0))**2 + tau_I**2)**0.5\n", - "sigma_I_2 = 1/2 * (sigma_I + 0) - ((1/2 * (sigma_I + 0))**2 + tau_I**2)**0.5\n", - "print(sigma_I_1, sigma_I_2)\n", - "\n", - "sigma_C = -8.53\n", - "tau_C = -tau_C\n", - "sigma_C_1 = 1/2 * (sigma_C + 0) + ((1/2 * (sigma_C))**2 + tau_C**2)**0.5\n", - "sigma_C_2 = 1/2 * (sigma_C + 0) - ((1/2 * (sigma_C))**2 + tau_C**2)**0.5\n", - "print(sigma_C_1, sigma_C_2)" - ] - }, - { - "cell_type": "code", - "execution_count": null, - "metadata": {}, - "outputs": [], - "source": [] - } - ], - "metadata": { - "kernelspec": { - "display_name": "base", - "language": "python", - "name": "python3" - }, - "language_info": { - "codemirror_mode": { - "name": "ipython", - "version": 3 - }, - "file_extension": ".py", - "mimetype": "text/x-python", - "name": "python", - "nbconvert_exporter": "python", - "pygments_lexer": "ipython3", - "version": "3.9.18" - } - }, - "nbformat": 4, - "nbformat_minor": 2 -} diff --git a/book/week_13/session_1/intro.py b/book/week_13/session_1/intro.py new file mode 100644 index 0000000..c5c2e60 --- /dev/null +++ b/book/week_13/session_1/intro.py @@ -0,0 +1,113 @@ +# --- +# jupyter: +# jupytext: +# cell_markers: '"""' +# text_representation: +# extension: .py +# format_name: percent +# format_version: '1.3' +# jupytext_version: 1.16.1 +# kernelspec: +# display_name: base +# language: python +# name: python3 +# --- + +# %% [markdown] +r""" +```{index} Transformations; Class exercise using analytical formulas +``` + +(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 +""" + +# %% 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)