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

Reactions

$\begin{aligned} &\sum{F_{iX}}=0 \\ & H_A=0 &\\ &\sum{M_{A}}=0 \\ & -V_B\cdot 10m+Q\cdot 5m=0 & V_B=50\ kN\\ &\sum{M_{B}}=0 \\ & V_A\cdot 10m-Q\cdot 5m=0 & V_A=50\ kN\\ &\sum{F_{iY}}=0 \\ & V_A+V_B-Q=0 & L=P\\ \end{aligned}$

\begin{aligned} M_{max}=V_A\cdot 5-\frac{1}{2}\cdot 10\cdot 5^2=125\ kN \end{aligned}

Stresses

\begin{aligned} I_Y=\frac{bh^3}{12}=\frac{10\cdot 30^3}{12}=22500\ cm^4 \end{aligned}

\begin{aligned} \sigma_d=\frac{M_Y}{I_Y}z_d \end{aligned}

\begin{aligned} \sigma_d=-\frac{125\cdot 10^3}{22500\cdot 10^{-8}}\cdot 15\cdot 10^{-2}=-83,33\ MPa \end{aligned}

Due to symmetry

\begin{aligned} \sigma_g=-\sigma_A=83,33\ MPa \end{aligned}

Graph

Bending

Strength of materials - Bending - Example 2

Example 2