Understanding the following will be aided by the video course from Example 1.

What you should know before starting this example: Geometric characteristics.

\begin{aligned} &\sum{M_{A}}=0\\ &8\cdot 4\cdot\ 2\cdot -V_{B}\cdot 7=0\\ &V_{B}=9.143\\ &\sum{M_{B}}=0\\ &V_{A}\cdot 7-8\cdot 4\cdot 5=0\\ &V_{A}=22.857\\ \end{aligned}

Geometric characteristics

\begin{aligned} &Z_{1}=9 \ m \\ &A_{1}=18\cdot 14=252 \ cm^{2}\\ &Z_{2}=10 \ m \\ &A_{2}=16\cdot 10=160 \ cm^{2}\\ &Z_{C}=\frac{Z_{1}A_{1}-Z_{2}A_{2}}{A_{1}-A_{2}}=7.261\\ &I_{yc}=I_{y_{1}}+A_{1}(Z_{1}-Z_{C})^{2}-[I_{y_{2}}+A_{2}(Z_{2}-Z_{C})^{2}]=2952.406 \ cm^{4}\\ \end{aligned}

Tangential stresses along the section height

\begin{aligned} & \tau_{xz}=\frac{F_{z}\cdot S_{y}(z)}{I_{z}\cdot b(z)}\\ \end{aligned} \begin{aligned} &S_{A}=0\\ \end{aligned}

\begin{aligned} &S_{B}=2\cdot 14\cdot 1=28 \ cm^{3}\\ \end{aligned}

\begin{aligned} &S_{C}=S_{B}+5.621\cdot 4\cdot\frac{5.621}{2}=91.19 \ cm^{3}\\ &S_{D}=0\\ \end{aligned} \begin{aligned} &\tau_{A}=0\\ &\tau_{B_g}=\frac{-9.143\cdot 10^{3}\cdot 28\cdot 10^{-6}}{2952.406\cdot 10^{-8}\cdot 14\cdot 10^{-6}}=-63.765 \ kPa\\ &\tau_{B_d}=-223.18 \ kPa\\ &\tau_{C}=\frac{-9.143\cdot 10^3\cdot 91.19\cdot 10^{-6}}{2952.406\cdot 10^{-8}\cdot 4\cdot 10^{-2}}=-726.84 \ kPa\\ &\tau_{D}=0\\ \end{aligned}

Tangential stresses along the section width

\begin{aligned} &\tau_{xy}=\frac{F_{z}\cdot S_{y}(y)}{I_{y}\cdot h(y)}\\ \end{aligned}

\begin{aligned} &S_{F}=0\\ &S_{G}=5\cdot 2\cdot 2.5=25 \ cm^{3}\\ &\tau_{xy}=\frac{-9.143\cdot 10^{3}\cdot 25\cdot 10^{-6}}{2952.406\cdot 10^{-8}\cdot 2\cdot 10^{-2}}=-387.099 kPa\\ \end{aligned}

Shear stress diagram

Bending

Strength of materials - Bending - Example 4

Example 4