test

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/

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)