```{index} Continuum mechanics; Exam assignment ``` ```{index} Buckling; Exam assignment ``` (exam3)= # Exam Friday January 30th Today you'll make the second exam assignment covering Continuuum mechanics including its prerequisites and/or the first exam assignment covering Buckling including its prerequisites. For more information about the exam see [the assessment information in course information](exam-general) ## Exam assignment 2 Continuum mechanics Your own submission and its grading will be available on [ANS exam assignment Continuum mechanics 2](https://ans.app/universities/1/courses/576319/assignments/1649807/go_to) after the exam. Given is the following structure: ::::{margin} :::{note} In the original exam, the $y$-axis was incorrectly pointing in the opposite direction. This does not affect the solution of the exercises. ::: :::: ```{figure} intro_data/constructie.svg :align: center :source: https://github.com/Structural-Mechanics-CEG/mechanics-figures-source/tree/main/shear_2 ``` :::::{exercise} :nonumber: true Show that $A = 78750 \, \rm{mm^2}$, $\bar z_{\rm{NC}} \approx 167 \, \rm{mm}$ and $I_{zz} = 1.1 \cdot 10^9 \, \rm{mm^4}$. ::::: ::::{admonition} Solution :class: solution, dropdown $$ A = 100 \cdot 75 \cdot 4 + 2 \cdot \frac{1}{2} \cdot 75 \cdot 50 + 2 \cdot 75 \cdot 50 + 250 \cdot 2 \cdot 75 = 78750 \, \rm{mm}^2 $$ $$ \begin{align*} \bar z_{\rm{N.C.}} = &\cfrac{S_{\bar z}}{A} \\ = &\cfrac{\frac{1}{2}\cdot 100 \cdot100 \cdot 75 \cdot 4}{78750} \\ &+ 2\cdot \cfrac{\left(100 + \frac{1}{3}\cdot 50\right) \cdot \frac{1}{2} \cdot 75 \cdot 50}{78750}\\ & + \cfrac{\left(100+\frac{1}{2} \cdot 50 \right) \cdot 2 \cdot 75 \cdot 50}{78750}\\ & + \cfrac{\left( 100 + 50 + \frac{1}{2}\cdot 250 \right) \cdot 250 \cdot 2 \cdot 75 }{78750} \\ \approx &167 \, \rm{mm} \end{align*} $$ $$ \begin{align*} I_{zz} = &\frac{1}{12} \cdot 4 \cdot 75 \cdot 100^3 + 4 \cdot 75 \cdot 100 \cdot \left( \frac{1}{2} \cdot 100 - 167 \right)^2 \\ &+ 2 \cdot \left( \frac{1}{36} \cdot 75 \cdot 50^3 + \frac{1}{2} \cdot 75 \cdot 50 \cdot \left( 100 + \frac{1}{3}\cdot 50 - 167 \right)^2\right) \\ &+ \frac{1}{12} \cdot 2 \cdot 75 \cdot 50^3 + 2 \cdot 75 \cdot 50 \cdot \left(100+\frac{1}{2} \cdot 50 - 167 \right)^2 \\ &+ \frac{1}{12} \cdot 2 \cdot 75 \cdot 250^3 + 250 \cdot 2 \cdot 75 \cdot \left( 100 + 50 + \frac{1}{2}\cdot 250 - 167 \right)^2 \\ \approx & 1093 \cdot10^6\, \rm{mm^4} \end{align*} $$ :::: :::::{exercise} :nonumber: true Show that $M_{\rm{C}} = 1800 \, \rm{kNm}$ (◡) and $V_{\rm{C}} = 600 \, \rm{kN}$ (⎽|⎺). ::::: ::::{admonition} Solution :class: solution, dropdown ```{figure} ./intro_data/VLS_1.svg :align: center :source: https://github.com/Tom-van-Woudenberg/mechanics-figures-source/tree/main/shear_2 ``` $$ \sum {T_{\rm{A}}} = 0 \to B_{\rm{v}} = 600 \, \rm{kN} (↑) $$ ```{figure} ./intro_data/VLS_2.svg :align: center :source: https://github.com/Tom-van-Woudenberg/mechanics-figures-source/tree/main/shear_2 ``` $$ \sum {F_{\rm{v}}^{\rm{CB}}} = 0 \to V_{\rm{C}} = 600 \, \rm{kN} (⎽|⎺) $$ $$ \sum {T_{\rm{C}}^{\rm{CB}}} = 0 \to M_{\rm{C}} = 1800 \, \rm{kNm} (◡) $$ :::: :::::{exercise} :nonumber: true Find the 3D principal stress tensor at point $\rm{D}$ in cross-section $\rm{C}$ and indicate the direction. ::::: ::::{admonition} Solution :class: solution, dropdown $$ S_{\rm{z}}^{\rm{a}} = 150 \cdot 75 \cdot 2 \cdot \left( 100+50+100+\frac{1}{2}\cdot150 - 167\right) \approx 3.54 \cdot 10^6 \, \rm{mm}^3$$ $$ \tau_{\rm{D}} = \cfrac{ 600000 \cdot 3.54 \cdot 10^6 }{ 1093 \cdot 10^6 \cdot 2 \cdot 75 } \approx 13 \, \rm{MPa} (↑) $$ $$ \sigma_{\rm{D}} = \cfrac{ 1800 \cdot 10^6 \cdot \left( 100 + 50 + 100 - 167 \right) }{ 1093 \cdot 10^6 } \approx +136 \, \rm{MPa} $$ This leads to the following stress tensor at point $\rm{D}$ in cross-section $\rm{C}$ according to the given coordinate system: $$ \boldsymbol{\sigma} = \begin{bmatrix} 136 & 0 & -13 \\ 0 & 0 & 0 \\ -13 & 0 & 0 \end{bmatrix} \, \rm{MPa} $$ In an upward $y$, rightward $x$ coordinate system, the stress tensor is: $$ \boldsymbol{\sigma} = \begin{bmatrix} 136 & 13 & 0 \\ 13 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \, \rm{MPa} $$ The direction of the principal stresses can be found using: $$ \alpha = \frac{1}{2} \cdot \arctan \left( \cfrac{13}{\frac{1}{2} \cdot\left(136 - 0\right)} \right) = 5.4^\circ $$ The principal stresses can be found using: $$ \begin{align*} \sigma_1 &= \frac{1}{2} \cdot \left( 136 - 0 \right) + \frac{1}{2} \cdot \left(136 - 0 \right) \cdot \cos \left( 2 \cdot 5.4 \right) + 13 \cdot \sin \left( 2 \cdot 5.4 \right) = 137 \, \rm{MPa} \\ \sigma_2 &= \frac{1}{2} \cdot \left( 136 - 0 \right) - \frac{1}{2} \cdot \left(136 - 0 \right) \cdot \cos \left( 2 \cdot 5.4 \right) - 13 \cdot\sin \left( 2 \cdot 5.4 \right) = -1.2 \, \rm{MPa} \\ \end{align*} $$ Leading to: $$ \boldsymbol{\bar \sigma} = \begin{bmatrix} 137 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1.2 \end{bmatrix} \, \rm{MPa} $$ In the following direction: ```{figure} ./intro_data/direction.svg :align: center :source: https://github.com/Tom-van-Woudenberg/mechanics-figures-source/tree/main/shear_2 ``` :::: ## Exam assignment 1 Buckling The exam assignment was provided as shown [here](https://github.com/Structural-Mechanics-CEG/mechanics-figures-source/raw/refs/heads/main/stab_exam/CT1000S-stab-deeltentamen-JAN2026.pdf). The solution is included in the same file.