diff --git a/book/_config.yml b/book/_config.yml index b42816d..53977b9 100644 --- a/book/_config.yml +++ b/book/_config.yml @@ -39,10 +39,6 @@ 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 new file mode 100644 index 0000000..690b583 --- /dev/null +++ b/book/week_13/session_1/intro.ipynb @@ -0,0 +1,154 @@ +{ + "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 deleted file mode 100644 index c5c2e60..0000000 --- a/book/week_13/session_1/intro.py +++ /dev/null @@ -1,113 +0,0 @@ -# --- -# 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)