Cross-sectional forces

\begin{aligned} &\sum x=0 \\ &H_{A}=60 k N \\ &\sum M_{A}=5+5 \cdot 5 \cdot 2.5-V_{B} \cdot 5=0 \\ &V_{B}=13.5 k N \\ &\sum y=0 \\ &V_{A}-5 \cdot 5+13.5=0 \\ &V_{A}=11.5 k N \end{aligned}

Bending moment function:

\begin{aligned} M(x)=5+11.5 x-\frac{1}{2} \cdot 5 \cdot x^{2} \end{aligned}

Checking for extremum:

\begin{aligned} &\frac{d M(x)}{d x}=11.5-5 x \\ &11.5-5 x=0 \Rightarrow x=2.3 \\ &M(0)=5 k N m \\ &M(5)=0 k N m \\ &M_{\text {extr }}(2.3)=5+11.5 \cdot 2.3-2.5 \cdot 2.3^{2}=18.225 k N m \end{aligned}

Normal force and bending moment graph

Normal stresses

\begin{aligned} &\sigma_{N}=\frac{N}{A} \\ &A=10 \cdot 50+10 \cdot 70=1200 \mathrm{~cm}^{2} \\ &\sigma_{N}=\frac{-60 \cdot 10^{3}}{1200 \cdot 10^{-4}}=-0.5 \mathrm{MPa} \\ &\sigma_{G}=\frac{M_{y}}{I_{y}} z \end{aligned}

Geometric characteristics

\begin{aligned} z_{1} &=25 \mathrm{~mm} \\ z_{2} &=55 \mathrm{~mm} \\ A_{1} &=500 \mathrm{~mm}^{2} \\ A_{2} &=700 \mathrm{~mm}^{2} \\ z_{c} &=\frac{z_{1} \cdot A_{1}+z_{2} \cdot A_{2}}{A_{1}+A_{2}}=42.5 \mathrm{~mm} \\ I_{y c} &=\frac{10 \cdot 50^{3}}{12}+500 \cdot(25-42.5)^{2}+\frac{70 \cdot 10^{3}}{12}+700 \cdot(55-42.5)^{2}=372500 \mathrm{~cm}^{4} \\ \sigma_{G}^{d} &=\frac{18.225 \cdot 10^{3}}{372500 \cdot 10^{-8}} \cdot 0.0425=0.21 \mathrm{MPa} \\ \sigma_{G}^{g} &=\frac{18.225 \cdot 10^{3}}{372500 \cdot 10^{-8}} \cdot(0.06-0.0425)=-0.09 \mathrm{MPa} \end{aligned} The final graph of normal stresses is obtained by adding compressive and tensile stresses at characteristic points (on the upper and lower fibers).