Geometric Characteristics

\begin{aligned} &A_1=20\cdot 80=1600\ mm^2 \\ &A_2=20\cdot 100=2000\ mm^2 \\ &z_1=40\ mm \\ &z_2=90\ mm \\ &z_c=\frac{Sy}{A}=\frac{z_1\cdot A_1+z_2\cdot A_2}{A_1+A_2} \\ &z_c=67.78\ mm\\ &I_{yc}=I_{y1}+A_1\cdot (z_1-z_c)^2+I_{y2}+A_2\cdot (z_2-z_c)^2\\ &I_{yc}=\frac{20\cdot 80^3}{12}+1600\cdot (40-67.78)^2+\frac{100\cdot 20^3}{12}+2000\cdot (90-67.78)^2=3083888.9\ mm^4=3.08\cdot 10^{-6}\ m^4\\ \end{aligned}

Normal Stresses

\begin{aligned} &\sigma =\frac{M}{I_{yc}}\cdot z\\ &\sigma_5 =\frac{5000}{3.08\cdot 10^{-6}}\cdot 32.22\cdot 10^{-3}=52.31\ MPa\\ &\sigma_1 =110.03\ MPa\\ \end{aligned}

Tangential Stresses

\begin{aligned} &S_1=S_5=0\\ \end{aligned}

\begin{aligned} &S_3=S_4=100\cdot 20\cdot 22.22=44440\ mm^3=4.44\cdot 10^{-5}\ m^3\\ \end{aligned}

\begin{aligned} &S_2=S_4+12.22\cdot 20\cdot 6.11=4.59\cdot 10^{-5}\ m^3\\ \end{aligned} \begin{aligned} &\tau=\frac{T\cdot S}{b\cdot I_{xc}}=\frac{100\cdot 10^3}{3.08\cdot 10^{-6}}\cdot \frac{S}{b}=3.25\cdot 10^{10}\cdot \frac{S}{b} &\tau_1=\tau_5=0\\ &\tau_4=3.25\cdot 10^{10}\cdot \frac{4.44\cdot 10^{-5}}{0.1}=14.43\ MPa\\ &\tau_3=3.25\cdot 10^{10}\cdot \frac{4.44\cdot 10^{-5}}{0.02}=72.15\ MPa\\ &\tau_2=3.25\cdot 10^{10}\cdot \frac{4.59\cdot 10^{-5}}{0.02}=74.59\ MPa\\ \end{aligned}

Normal and Tangential Stress Chart

Bending

Strength of materials - Bending - Example 8

Example 8