typo fixes

book/week_10/session/intro.md

@@ -29,23 +29,23 @@ For example when using the force method:
 ```
 
 This gives:
-- ${N_{{\rm{AD}}}} = - \cfrac{4}{5}{B_{\rm{v}}}$
-- ${N_{{\rm{BD}}}} = + {B_{\rm{v}}}$
-- ${N_{{\rm{CD}}}} = - \cfrac{3}{4}{B_{\rm{v}}}$
+- ${N_{{\text{AD}}}} = - \cfrac{4}{5}{B_{\text{v}}}$
+- ${N_{{\text{BD}}}} = + {B_{\text{v}}}$
+- ${N_{{\text{CD}}}} = - \cfrac{3}{4}{B_{\text{v}}}$
 
 This gives elongations:
 
-- $\Delta {L_{{\rm{BD}}}} = \cfrac{{{B_{\rm{v}}}}}{{7500}}$
-- $\Delta {L_{{\rm{AD}}}} = - \cfrac{{{B_{\rm{v}}}}}{{4800}}$
-- $\Delta {L_{{\rm{CD}}}} = - \cfrac{{{B_{\rm{v}}}}}{{10000}}$
+- $\Delta {L_{{\text{BD}}}} = \cfrac{{{B_{\text{v}}}}}{{7500}}$
+- $\Delta {L_{{\text{AD}}}} = - \cfrac{{{B_{\text{v}}}}}{{4800}}$
+- $\Delta {L_{{\text{CD}}}} = - \cfrac{{{B_{\text{v}}}}}{{10000}}$
 
 Resulting in a williot like this (keeping the unknown value of $$B_\text{v}$$ constant for all elongations):
 
 ```{figure} intro_data/williot.svg
 :align: center
 ```
 
-This gives a vertical displacement of $\text{B}$ of $\cfrac{{3{B_{\rm{v}}}}}{{6400}}$.
+This gives a vertical displacement of $\text{B}$ of $\cfrac{{3{B_{\text{v}}}}}{{6400}}$.
 
 With the requirements of $30 \text{ mm}$ this leads to $B_\text{v} = 64 \text{ kN}$, resulting in:
 

book/week_11/session_1/intro.ipynb

@@ -101,7 +101,7 @@
     "$$\\sigma  = \\left[ \\begin{array}{}\n",
     "{15}&{-6.9}\\\\\n",
     "{-6.9}&{0}\n",
-    "\\end{array} \\right]{\\rm{ MPa}}$$\n",
+    "\\end{array} \\right]{\\text{ MPa}}$$\n",
     "\n",
     "The maximum stress can be found by applying the transformation rules:\n",
     "\n",

book/week_13/session_3/intro.ipynb

@@ -43,7 +43,7 @@
     "````{admonition} Solution assignment 2\n",
     ":class: tip, dropdown\n",
     "\n",
-    "$\\varphi_{\\text{E},x} = -1.1 \\cdot 10^{-4} \\text{ rad}$\n",
+    "$\\varphi_{\\text{E,}x} = -1.1 \\cdot 10^{-4} \\text{ rad}$\n",
     "\n",
     "````\n",
     "\n",
@@ -68,7 +68,7 @@
     "````{admonition} Solution assignment 4\n",
     ":class: tip, dropdown\n",
     "\n",
-    "$\\varphi_{\\text{B},x} = -0.0254 \\text{ rad}$\n",
+    "$\\varphi_{\\text{B,}x} = -0.0254 \\text{ rad}$\n",
     "\n",
     "````"
    ]

book/week_2/session_1/intro.ipynb

@@ -73,7 +73,7 @@
     "\n",
     "```{math}\n",
     ":label: BC_2\n",
-    "\\sum {{{\\left. {{F_{\\rm{h}}}} \\right|}^{\\rm{A}}} = 0}  \\to \\frac{4}{5} \\cdot {N_{{\\rm{AC}}}} + 150 = 0\n",
+    "\\sum {{{\\left. {{F_{\\text{h}}}} \\right|}^{\\text{A}}} = 0}  \\to \\frac{4}{5} \\cdot {N_{{\\text{AC}}}} + 150 = 0\n",
     "```\n",
     "\n",
     "This element could be solved independently as there are two unknowns (two integration constants) and two equations ({eq}`BC_1` and {eq}`BC_2`)."
@@ -114,7 +114,7 @@
     "\n",
     "```{math}\n",
     ":label: BC_5\n",
-    "{u_{{\\rm{AD}}}}\\left( 0 \\right) = {w_{{\\rm{DE}}}}\\left( 0 \\right)\n",
+    "{u_{{\\text{AD}}}}\\left( 0 \\right) = {w_{{\\text{DE}}}}\\left( 0 \\right)\n",
     "```\n",
     "\n",
     "And finally, a boundary condition can be formulated as the normal force in $\\text{D}$ is in equilibrium with the forces from element $\\text{DE}$. The free-body-diagram of $\\text{D}$ is:\n",
@@ -129,7 +129,7 @@
     "\n",
     "```{math}\n",
     ":label: BC_6\n",
-    "\\sum {{{\\left. {{F_{\\rm{v}}}} \\right|}^{\\rm{D}}} = 0}  \\to {N_{{\\rm{AD}}}} + V_{\\rm{D}}^{{\\rm{DE}}} = 0\n",
+    "\\sum {{{\\left. {{F_{\\text{v}}}} \\right|}^{\\text{D}}} = 0}  \\to {N_{{\\text{AD}}}} + V_{\\text{D}}^{{\\text{DE}}} = 0\n",
     "```"
    ]
   },
@@ -194,14 +194,14 @@
     "\n",
     "```{math}\n",
     ":label: BC_10\n",
-    "\\sum {{{\\left. {{F_{\\rm{v}}}} \\right|}^{\\rm{E}}} = 0}  \\to -{V_{{\\rm{E}}^\\text{DE}}} + V_{\\rm{E}}^{{\\rm{BE}}} = 0\n",
+    "\\sum {{{\\left. {{F_{\\text{v}}}} \\right|}^{\\text{E}}} = 0}  \\to -{V_{{\\text{E}}^\\text{DE}}} + V_{\\text{E}}^{{\\text{BE}}} = 0\n",
     "```\n",
     "\n",
     "Moment equilibrium gives:\n",
     "\n",
     "```{math}\n",
     ":label: BC_11\n",
-    "\\sum {{{\\left. {{T}} \\right|}_\\text{E}^{\\rm{E}}} = 0}  \\to -{M_{{\\rm{E}}^\\text{DE}}} + M_{\\rm{E}}^{{\\rm{BE}}} = 0\n",
+    "\\sum {{{\\left. {{T}} \\right|}_\\text{E}^{\\text{E}}} = 0}  \\to -{M_{{\\text{E}}^\\text{DE}}} + M_{\\text{E}}^{{\\text{BE}}} = 0\n",
     "```"
    ]
   },

book/week_2/session_3/intro.ipynb

@@ -44,7 +44,7 @@
     "````{admonition} Solution assignment 1 and 2\n",
     ":class: tip, dropdown\n",
     "\n",
-    "${w_{\\rm{C}}} =\\cfrac {1}{18} \\approx {\\rm{0}}{\\rm{.0556}}{\\text{ m}}$ ↓\n",
+    "${w_{\\text{C}}} =\\cfrac {1}{18} \\approx {\\text{0}}{\\text{.0556}}{\\text{ m}}$ ↓\n",
     "````\n",
     "\n",
     "````{admonition} Solution assignment 3\n",

book/week_3/session_3/intro.md

@@ -50,9 +50,9 @@ Free-body-diagram full structure
 
 $$
 \begin{array}{c}
-\sum {{{\left. T \right|}_{\rm{A}}} = 0}  \to {B_{\rm{v}}} = 5{\text{ kN}}\\
-\sum {{F_{\rm{v}}} = 0}  \to {A_{\rm{v}}} = 15{\text{ kN}}\\
-\sum {{F_{\rm{h}}} = 0}  \to {A_{\rm{h}}} = {B_{\rm{h}}}
+\sum {{{\left. T \right|}_{\text{A}}} = 0}  \to {B_{\text{v}}} = 5{\text{ kN}}\\
+\sum {{F_{\text{v}}} = 0}  \to {A_{\text{v}}} = 15{\text{ kN}}\\
+\sum {{F_{\text{h}}} = 0}  \to {A_{\text{h}}} = {B_{\text{h}}}
 \end{array}
 $$
 
@@ -76,8 +76,8 @@ Free-body-diagram joint $\text{B}$
 
 $$
 \begin{array}{c}
-\sum {{F_{\rm{v}}} = 0}  \to {N_{{\rm{BE}}}} = -6.25{\text{ kN}}\\
-\sum {{F_{\rm{h}}} = 0}  \to {N_{{\rm{BD}}}} =  3.75 - {B_{\rm{h}}}
+\sum {{F_{\text{v}}} = 0}  \to {N_{{\text{BE}}}} = -6.25{\text{ kN}}\\
+\sum {{F_{\text{h}}} = 0}  \to {N_{{\text{BD}}}} =  3.75 - {B_{\text{h}}}
 \end{array}
 $$
 
@@ -97,9 +97,9 @@ Free-body-diagram part $\text{AC}$
 
 $$
 \begin{array}{c}
-\sum {{F_{\rm{v}}} = 0}  \to {N_{{\rm{CD}}}} =  - 6.25{\rm{ kN}}\\
-{\sum {\left. T \right|} _{\rm{D}}} = 0 \to {N_{CE}} =  - 7.5{\rm{ kN}}\\
-\sum {{F_{\rm{h}}} = 0}  \to {N_{{\rm{AD}}}} = 11.25 - {B_{\rm{h}}}
+\sum {{F_{\text{v}}} = 0}  \to {N_{{\text{CD}}}} =  - 6.25{\text{ kN}}\\
+{\sum {\left. T \right|} _{\text{D}}} = 0 \to {N_{CE}} =  - 7.5{\text{ kN}}\\
+\sum {{F_{\text{h}}} = 0}  \to {N_{{\text{AD}}}} = 11.25 - {B_{\text{h}}}
 \end{array}
 $$
 
@@ -117,7 +117,7 @@ Thirdly, let's continue with the joint $\text{D}$:
 Free-body-diagram joint $\text{D}$
 ```
 
-$$\sum {{F_{\rm{v}}} = 0}  \to {N_{{\rm{DE}}}} =  6.25{\text{ kN}}$$
+$$\sum {{F_{\text{v}}} = 0}  \to {N_{{\text{DE}}}} =  6.25{\text{ kN}}$$
 
 ```{figure} ./intro_data/FBD_D_sol.svg
 :align: center
@@ -133,7 +133,7 @@ And finally joint $\text{C}$:
 Free-body-diagram joint $\text{D}$
 ```
 
-$$\sum {{F_{\rm{v}}} = 0}  \to {N_{{\rm{AC}}}} =  - 18.75{\text{ kN}}$$
+$$\sum {{F_{\text{v}}} = 0}  \to {N_{{\text{AC}}}} =  - 18.75{\text{ kN}}$$
 
 ```{figure} ./intro_data/FBD_C_sol.svg
 :align: center
@@ -146,13 +146,13 @@ Free-body-diagram joint $\text{D}$
 Now, for each element the shortening / lengthening can be calculated:
 
 $$\Delta L = \frac{{NL}}{{EA}} \to \begin{array}{c}
-{\Delta {L_{{\rm{AC}}}} =  - 0.025{\text{ m}}}\\
-{\Delta {L_{{\rm{CE}}}} =  - 0.012{\text{ m}}}\\
-{\Delta {L_{\rm{BE}}} = \cfrac{1}{{120}} \approx  - 0.00833{\text{ m}}}\\
-{\Delta {L_{{\rm{CD}}}} = \cfrac{1}{{120}} \approx  - 0.00833{\text{ m}}}\\
-{\Delta {L_{{\rm{DE}}}} = \cfrac{1}{{120}} \approx 0.00833{\text{ m}}}\\
-{\Delta {L_{{\rm{AD}}}} = 0.018 - \cfrac{1}{{625}}{B_{\rm{h}}} = 0.018 - 0.0016{B_{\rm{h}}}{\text{ m}}}\\
-{\Delta {L_{{\rm{DB}}}} = 0.006 - 0.0016{B_{\rm{h}}}{\text{ m}}}
+{\Delta {L_{{\text{AC}}}} =  - 0.025{\text{ m}}}\\
+{\Delta {L_{{\text{CE}}}} =  - 0.012{\text{ m}}}\\
+{\Delta {L_{\text{BE}}} = \cfrac{1}{{120}} \approx  - 0.00833{\text{ m}}}\\
+{\Delta {L_{{\text{CD}}}} = \cfrac{1}{{120}} \approx  - 0.00833{\text{ m}}}\\
+{\Delta {L_{{\text{DE}}}} = \cfrac{1}{{120}} \approx 0.00833{\text{ m}}}\\
+{\Delta {L_{{\text{AD}}}} = 0.018 - \cfrac{1}{{625}}{B_{\text{h}}} = 0.018 - 0.0016{B_{\text{h}}}{\text{ m}}}\\
+{\Delta {L_{{\text{DB}}}} = 0.006 - 0.0016{B_{\text{h}}}{\text{ m}}}
 \end{array}$$
 
 ### Displacement structure due to $20 \text{ kN}$
@@ -223,12 +223,12 @@ Leading to the following displacements if $\text{AD}$ doesn't rotate:
 | joint | Displacement due to $B_\text{h}$ in horizontal direction → | Displacement due to $B_\text{h}$ in vertical direction ↓|
 | :-:|:-:|:-:|
 |$\text{A}$|$0$|$0$|
-|$\text{C}$|$-0.8{B_{\rm{h}}}$|$-0.6{B_{\rm{h}}}$|
-|$\text{D}$|$-1.6{B_{\rm{h}}}$|$0$|
-|$\text{E}$|$-0.8{B_{\rm{h}}}$|$0.6{B_{\rm{h}}}$|
-|$\text{B}$|$-3.2{B_{\rm{h}}}$|$2.4{B_{\rm{h}}}$|
+|$\text{C}$|$-0.8{B_{\text{h}}}$|$-0.6{B_{\text{h}}}$|
+|$\text{D}$|$-1.6{B_{\text{h}}}$|$0$|
+|$\text{E}$|$-0.8{B_{\text{h}}}$|$0.6{B_{\text{h}}}$|
+|$\text{B}$|$-3.2{B_{\text{h}}}$|$2.4{B_{\text{h}}}$|
 
-Again, $\text{B}$ Shouldn't move vertically, so this structure has to be rotated back with $\theta = \cfrac{2.4{B_{\rm{h}}}}{{12000}} = 0.0002{B_{\rm{h}}}{\text{ rad}}$ ⟲, leading to: 
+Again, $\text{B}$ Shouldn't move vertically, so this structure has to be rotated back with $\theta = \cfrac{2.4{B_{\text{h}}}}{{12000}} = 0.0002{B_{\text{h}}}{\text{ rad}}$ ⟲, leading to: 
 
 | joint | Displacement due to $\theta$ in horizontal direction → $\left( \text{mm}\right)$| Displacement due to $\theta$ in vertical direction ↓ $\left( \text{mm}\right)$|
 | :-:|:-:|:-:|
@@ -244,7 +244,7 @@ Resulting in total displacements of:
 | :-:|:-:|:-:|
 |$\text{A}$|$0$|$0$|
 |$\text{C}$|$-1.6B_\text{h}$|$-1.2B_\text{h}$|
-|$\text{D}$|$-1.6{B_{\rm{h}}}$|$-1.2B_\text{h}$|
+|$\text{D}$|$-1.6{B_{\text{h}}}$|$-1.2B_\text{h}$|
 |$\text{E}$|$-1.6B_\text{h}$|$-1.2B_\text{h}$|
 |$\text{B}$|$-3.2B_\text{h}$|$0$|
 
@@ -258,7 +258,7 @@ Displaced structure
 
 Now, we can fill in the compatibility conditions:
 
-$${u_{{\rm{B,h}}}} = 0 \to 0.024 - 0.0032{B_{\rm{h}}} = 0 \to {B_{\text{h}}} =  7.5{\text{ kN}}$$
+$${u_{{\text{B,h}}}} = 0 \to 0.024 - 0.0032{B_{\text{h}}} = 0 \to {B_{\text{h}}} =  7.5{\text{ kN}}$$
 
 ### Section forces statically indeterminate structure
 

book/week_5/session_1/intro.ipynb

@@ -25,7 +25,7 @@
     "\n",
     "### Define mechanism compatibility conditions\n",
     "\n",
-    "To apply the 'hoekveranderingsvergelijkingen', we apply hinges, with statically indeterminate moments $M_\\text{B}$ and $M_\\text{D}$ and compatibility conditions $\\varphi _{\\rm{B}}^{{\\rm{AB}}} = \\varphi _{\\rm{B}}^{{\\rm{BD}}} $ and $\\varphi _{\\rm{D}}^{{\\rm{BD}}} = \\varphi _{\\rm{D}}^{{\\rm{DC}}}$\n",
+    "To apply the 'hoekveranderingsvergelijkingen', we apply hinges, with statically indeterminate moments $M_\\text{B}$ and $M_\\text{D}$ and compatibility conditions $\\varphi _{\\text{B}}^{{\\text{AB}}} = \\varphi _{\\text{B}}^{{\\text{BD}}} $ and $\\varphi _{\\text{D}}^{{\\text{BD}}} = \\varphi _{\\text{D}}^{{\\text{DC}}}$\n",
     "\n",
     "```{figure} ./intro_data/statically_determinate.svg\n",
     ":align: center\n",
@@ -51,10 +51,10 @@
     "\n",
     "These rotations can be found using the forget-me-nots and the rotations of each of the elements due to the rigid body rotations:\n",
     "\n",
-    "- $ \\varphi _{\\rm{B}}^{{\\rm{AB}}} = \\cfrac{M_\\text{B}\\cdot 4}{3\\cdot 20000} \\to \\varphi _{\\rm{B}}^{{\\rm{AB}}} = \\cfrac{M_\\text{B}\\cdot 2}{30000} $\n",
-    "- $ \\varphi _{\\rm{B}}^{{\\rm{BD}}} = -\\cfrac{M_\\text{B}\\cdot 4}{3\\cdot 20000} - \\cfrac{M_\\text{D}\\cdot 4}{6\\cdot 20000} - \\cfrac{20 \\cdot 4^3}{24 \\cdot20000} - \\theta \\to \\varphi _{\\rm{B}}^{{\\rm{BD}}} = -\\cfrac{M_\\text{B}\\cdot 2}{30000} - \\cfrac{M_\\text{D}}{30000} - \\cfrac{1}{375} - \\theta$\n",
-    "- $ \\varphi _{\\rm{D}}^{{\\rm{BD}}} = \\cfrac{M_\\text{B}\\cdot 4}{6\\cdot 20000} + \\cfrac{M_\\text{D}\\cdot 4}{3\\cdot 20000} + \\cfrac{20 \\cdot 4^3}{24 \\cdot20000} - \\theta \\to \\varphi _{\\rm{D}}^{{\\rm{BD}}} = \\cfrac{M_\\text{B}}{30000} + \\cfrac{M_\\text{D}\\cdot 2}{30000} + \\cfrac{1}{375} - \\theta$\n",
-    "- $ \\varphi _{\\rm{D}}^{{\\rm{DC}}} = - \\cfrac{M_\\text{D}\\cdot 4 \\sqrt{2}}{3\\cdot 20000 \\sqrt{2}} + \\theta \\to \\varphi _{\\rm{D}}^{{\\rm{DC}}} = -\\cfrac{M_\\text{D}\\cdot 2}{30000} + \\theta$\n",
+    "- $ \\varphi _{\\text{B}}^{{\\text{AB}}} = \\cfrac{M_\\text{B}\\cdot 4}{3\\cdot 20000} \\to \\varphi _{\\text{B}}^{{\\text{AB}}} = \\cfrac{M_\\text{B}\\cdot 2}{30000} $\n",
+    "- $ \\varphi _{\\text{B}}^{{\\text{BD}}} = -\\cfrac{M_\\text{B}\\cdot 4}{3\\cdot 20000} - \\cfrac{M_\\text{D}\\cdot 4}{6\\cdot 20000} - \\cfrac{20 \\cdot 4^3}{24 \\cdot20000} - \\theta \\to \\varphi _{\\text{B}}^{{\\text{BD}}} = -\\cfrac{M_\\text{B}\\cdot 2}{30000} - \\cfrac{M_\\text{D}}{30000} - \\cfrac{1}{375} - \\theta$\n",
+    "- $ \\varphi _{\\text{D}}^{{\\text{BD}}} = \\cfrac{M_\\text{B}\\cdot 4}{6\\cdot 20000} + \\cfrac{M_\\text{D}\\cdot 4}{3\\cdot 20000} + \\cfrac{20 \\cdot 4^3}{24 \\cdot20000} - \\theta \\to \\varphi _{\\text{D}}^{{\\text{BD}}} = \\cfrac{M_\\text{B}}{30000} + \\cfrac{M_\\text{D}\\cdot 2}{30000} + \\cfrac{1}{375} - \\theta$\n",
+    "- $ \\varphi _{\\text{D}}^{{\\text{DC}}} = - \\cfrac{M_\\text{D}\\cdot 4 \\sqrt{2}}{3\\cdot 20000 \\sqrt{2}} + \\theta \\to \\varphi _{\\text{D}}^{{\\text{DC}}} = -\\cfrac{M_\\text{D}\\cdot 2}{30000} + \\theta$\n",
     "\n",
     "### Solve statically indeterminate structure with compatibility conditions and equilibrium\n",
     "\n",
@@ -66,12 +66,12 @@
     "Mechanism for virtual work calculation\n",
     "```\n",
     "\n",
-    "This leads to the virtual work equations: $\\delta A = 0 \\to {M_{\\rm{B}}} \\cdot \\delta \\theta  + 20 \\cdot 4 \\cdot 2\\delta \\theta  - {M_{\\rm{D}}} \\cdot \\delta \\theta  - {M_{\\rm{D}}} \\cdot \\delta \\theta  = 0 \\to {M_{\\rm{B}}} + 160 - 2{M_{\\rm{D}}} = 0$\n",
+    "This leads to the virtual work equations: $\\delta A = 0 \\to {M_{\\text{B}}} \\cdot \\delta \\theta  + 20 \\cdot 4 \\cdot 2\\delta \\theta  - {M_{\\text{D}}} \\cdot \\delta \\theta  - {M_{\\text{D}}} \\cdot \\delta \\theta  = 0 \\to {M_{\\text{B}}} + 160 - 2{M_{\\text{D}}} = 0$\n",
     "\n",
     "The compatibility conditions are as follows:\n",
     "\n",
-    "- $\\varphi _{\\rm{B}}^{{\\rm{AB}}} = \\varphi _{\\rm{B}}^{{\\rm{BD}}} \\to \\cfrac{M_\\text{B}\\cdot 2}{30000} = -\\cfrac{M_\\text{B}\\cdot 2}{30000} - \\cfrac{M_\\text{D}}{30000} - \\cfrac{1}{375} - \\theta \\to \\cfrac{M_\\text{B}\\cdot 4}{30000} + \\cfrac{M_\\text{D}}{30000} + \\cfrac{1}{375} + \\theta = 0$\n",
-    "- $\\varphi _{\\rm{D}}^{{\\rm{BD}}} = \\varphi _{\\rm{D}}^{{\\rm{DC}}} \\to \\cfrac{M_\\text{B}}{30000} + \\cfrac{M_\\text{D}\\cdot 2}{30000} + \\cfrac{1}{375} - \\theta = -\\cfrac{M_\\text{D}\\cdot 2}{30000} + \\theta \\to \\cfrac{M_\\text{B}}{30000} + \\cfrac{M_\\text{D}\\cdot 4}{30000} + \\cfrac{1}{375} - 2\\theta = 0$\n",
+    "- $\\varphi _{\\text{B}}^{{\\text{AB}}} = \\varphi _{\\text{B}}^{{\\text{BD}}} \\to \\cfrac{M_\\text{B}\\cdot 2}{30000} = -\\cfrac{M_\\text{B}\\cdot 2}{30000} - \\cfrac{M_\\text{D}}{30000} - \\cfrac{1}{375} - \\theta \\to \\cfrac{M_\\text{B}\\cdot 4}{30000} + \\cfrac{M_\\text{D}}{30000} + \\cfrac{1}{375} + \\theta = 0$\n",
+    "- $\\varphi _{\\text{D}}^{{\\text{BD}}} = \\varphi _{\\text{D}}^{{\\text{DC}}} \\to \\cfrac{M_\\text{B}}{30000} + \\cfrac{M_\\text{D}\\cdot 2}{30000} + \\cfrac{1}{375} - \\theta = -\\cfrac{M_\\text{D}\\cdot 2}{30000} + \\theta \\to \\cfrac{M_\\text{B}}{30000} + \\cfrac{M_\\text{D}\\cdot 4}{30000} + \\cfrac{1}{375} - 2\\theta = 0$\n",
     "\n",
     "Solving these 3 equations gives:\n",
     "\n",

book/week_7/session_3/intro.ipynb

@@ -109,8 +109,8 @@
     "The first displacement condition can be filled in using forget-me-nots. The support settlement is taken into account as a (real) displacement and thus rotation:\n",
     "\n",
     "$$\\begin{array}{c}\n",
-    "\\varphi _{\\rm{B}}^{{\\rm{AB}}} = \\varphi _{\\rm{B}}^{{\\rm{BC}}}\\\\\n",
-    "\\cfrac{{ - {M_{\\rm{B}}} \\cdot 4}}{{3 \\cdot EI}} + \\cfrac{{17 \\cdot {4^3}}}{{24 \\cdot EI}} - \\cfrac{{{w_{\\rm{B}}}}}{4} = \\cfrac{{{M_{\\rm{B}}} \\cdot 6}}{{3 \\cdot EI}} - \\cfrac{{{M_{\\rm{C}}} \\cdot 6}}{{6 \\cdot EI}} + \\cfrac{{{w_{\\rm{B}}}}}{6}\n",
+    "\\varphi _{\\text{B}}^{{\\text{AB}}} = \\varphi _{\\text{B}}^{{\\text{BC}}}\\\\\n",
+    "\\cfrac{{ - {M_{\\text{B}}} \\cdot 4}}{{3 \\cdot EI}} + \\cfrac{{17 \\cdot {4^3}}}{{24 \\cdot EI}} - \\cfrac{{{w_{\\text{B}}}}}{4} = \\cfrac{{{M_{\\text{B}}} \\cdot 6}}{{3 \\cdot EI}} - \\cfrac{{{M_{\\text{C}}} \\cdot 6}}{{6 \\cdot EI}} + \\cfrac{{{w_{\\text{B}}}}}{6}\n",
     "\\end{array}$$\n",
     "\n",
     "The second displacement condition gives similarly:\n",