Solution
\[
A = 100 \cdot 75 \cdot 4 + 2 \cdot \frac{1}{2} \cdot 75 \cdot 50 + 2 \cdot 75 \cdot 50 + 250 \cdot 2 \cdot 75 = 78750 \, \rm{mm}^2
\]
\[\begin{split}
\begin{align*}
\bar z_{\rm{N.C.}} = &\cfrac{S_{\bar z}}{A} \\
= &\cfrac{\frac{1}{2}\cdot 100 \cdot100 \cdot 75 \cdot 4}{78750} \\
&+ 2\cdot \cfrac{\left(100 + \frac{1}{3}\cdot 50\right) \cdot \frac{1}{2} \cdot 75 \cdot 50}{78750}\\
& + \cfrac{\left(100+\frac{1}{2} \cdot 50 \right) \cdot 2 \cdot 75 \cdot 50}{78750}\\
& + \cfrac{\left( 100 + 50 + \frac{1}{2}\cdot 250 \right) \cdot 250 \cdot 2 \cdot 75 }{78750} \\
\approx &167 \, \rm{mm}
\end{align*}
\end{split}\]
\[\begin{split}
\begin{align*}
I_{zz} = &\frac{1}{12} \cdot 4 \cdot 75 \cdot 100^3 + 4 \cdot 75 \cdot 100 \cdot \left( \frac{1}{2} \cdot 100 - 167 \right)^2 \\
&+ 2 \cdot \left( \frac{1}{36} \cdot 75 \cdot 50^3 + \frac{1}{2} \cdot 75 \cdot 50 \cdot \left( 100 + \frac{1}{3}\cdot 50 - 167 \right)^2\right) \\
&+ \frac{1}{12} \cdot 2 \cdot 75 \cdot 50^3 + 2 \cdot 75 \cdot 50 \cdot \left(100+\frac{1}{2} \cdot 50 - 167 \right)^2 \\
&+ \frac{1}{12} \cdot 2 \cdot 75 \cdot 250^3 + 250 \cdot 2 \cdot 75 \cdot \left( 100 + 50 + \frac{1}{2}\cdot 250 - 167 \right)^2 \\
\approx & 1093 \cdot10^6\, \rm{mm^4}
\end{align*}
\end{split}\]
Solution
\[
\sum {T_{\rm{A}}} = 0 \to B_{\rm{v}} = 600 \, \rm{kN} (↑)
\]
\[
\sum {F_{\rm{v}}^{\rm{CB}}} = 0 \to V_{\rm{C}} = 600 \, \rm{kN} (⎽|⎺)
\]
\[
\sum {T_{\rm{C}}^{\rm{CB}}} = 0 \to M_{\rm{C}} = 1800 \, \rm{kNm} (â—¡)
\]
Solution
\[ S_{\rm{z}}^{\rm{a}} = 150 \cdot 75 \cdot 2 \cdot \left( 100+50+100+\frac{1}{2}\cdot150 - 167\right) \approx 3.54 \cdot 10^6 \, \rm{mm}^3\]
\[ \tau_{\rm{D}} = \cfrac{ 600000 \cdot 3.54 \cdot 10^6 }{ 1093 \cdot 10^6 \cdot 2 \cdot 75 } \approx 13 \, \rm{MPa} (↑) \]
\[ \sigma_{\rm{D}} = \cfrac{ 1800 \cdot 10^6 \cdot \left( 100 + 50 + 100 - 167 \right) }{ 1093 \cdot 10^6 } \approx +136 \, \rm{MPa} \]
This leads to the following stress tensor at point \(\rm{D}\) in cross-section \(\rm{C}\) according to the given coordinate system:
\[\begin{split}
\boldsymbol{\sigma} =
\begin{bmatrix}
136 & 0 & -13 \\
0 & 0 & 0 \\
-13 & 0 & 0
\end{bmatrix} \, \rm{MPa}
\end{split}\]
In an upward \(y\) , rightward \(x\) coordinate system, the stress tensor is:
\[\begin{split}
\boldsymbol{\sigma} =
\begin{bmatrix}
136 & 13 & 0 \\
13 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix} \, \rm{MPa}
\end{split}\]
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:
\[\begin{split}
\begin{align*}
\sigma_1 &= \frac{1}{2} \cdot \left( 136 - 0 \right) + \frac{1}{2} \cdot \left(136 - 0 \right) \cdot \cos \left( 2 \cdot 5.4 \right) + 13 \cdot \sin \left( 2 \cdot 5.4 \right) = 137 \, \rm{MPa} \\
\sigma_2 &= \frac{1}{2} \cdot \left( 136 - 0 \right) - \frac{1}{2} \cdot \left(136 - 0 \right) \cdot \cos \left( 2 \cdot 5.4 \right) - 13 \cdot\sin \left( 2 \cdot 5.4 \right) = -1.2 \, \rm{MPa} \\
\end{align*}
\end{split}\]
Leading to:
\[\begin{split}
\boldsymbol{\bar \sigma} =
\begin{bmatrix}
137 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1.2
\end{bmatrix} \, \rm{MPa}
\end{split}\]
In the following direction:
exam assignment Continuum mechanics 2 after the exam.
