What you should know before starting this example: charts for Simple beams.

Understanding this example will be aided by video courses from Example 1 and Example 2.

$\begin{aligned} &\sum{M_B}=0\\ &6-8\cdot 3\cdot 4,5+R_{Ay}\cdot 6=0\\ &R_{Ay}=17\\ &\sum{M_A}=0\\ &6+8\cdot 3\cdot 1,3-GR_{By}=0\\ &R_{By}=7\\ &\sum{y}=0\\ &R_{Ay}-8\cdot 3+R_{By}=0\\ &L=P &\frac{17}{x}=\frac{7}{3-x}\\ &x=2,125\\ \end{aligned}$

Cross-section characteristics

$\begin{aligned} &Z_{1}=2\ cm\\ &A_{1}=4\cdot 8=32\ cm^2\\ &Z_{2}=4-\frac{1}{3}\cdot 2=\frac{10}{3}\ cm\\ &A_{2}=\frac{1}{2}\cdot 6\cdot 2=6\ cm^2\\ &Z_{C}=\frac{Z_{1}A_{1}-Z_{2}A_{2}}{A_{1}-A_{2}}=\frac{22}{13}\ cm\\ &I_{yc}=I_{y1}+A_{1}(Z_{1}-Z_{C})^2-[I_{y2}+A_{2}(Z_{2}-Z_{C})^2]=\frac{8\cdot4^3}{12}+32(2-\frac{22}{13})^2-\frac{6\cdot 2^3}{36}-6(\frac{10}{3}-\frac{22}{13})^2=28,205\ cm^4 \end{aligned}$

Normal stresses

$\begin{aligned} &\sigma=\frac{M_{max}}{I_{yc}}Z\\ &\sigma_g=-\frac{21\cdot10^3}{28,205\cdot 10^{-8}}\cdot (4-\frac{22}{13})\cdot 10^{-2}=-1718,18\ MPa\\ &\sigma_d=-\frac{21\cdot 10^3}{28,205\cdot 10^{-8}}\cdot -\frac{22}{13}10^{-2}=1260\ MPa\\ \end{aligned}$

Normal stress chart

Bending

Strength of materials - Bending - Example 11

Example 11