Understanding the following will help in the video course with Example 1.

The moment vector is perpendicular to the plane of loading (see the diagram below).

The value of the moment at the most loaded section is 5.5 kNm.

This moment is decomposed into principal central directions.

We can immediately determine the principal central axes because, first of all, we know where the center of gravity is, and secondly, the figure has a symmetry axis (at a 45-degree angle).

\begin{aligned} &I_{zc}=\frac{9\sqrt{2}\cdot (4.5\sqrt{2})^3}{36}=91.125\ cm^4\\ &I_{yc}=\frac{4.5\sqrt{2}\cdot (9\sqrt{2})^3}{48}=273.375\ cm^4\\ \\ &M=5.5\ kNm\\ &M_{yc}=M_{zc}=M\cdot \frac{\sqrt{2}}{2}=3.89\ kNm\\ \\ &\sigma=\frac{M_{yc}}{I_{yc}}zc-\frac{M_{zc}}{I_{zc}}yc\\ \end{aligned}

Neutral axis

\begin{aligned} &\frac{3.89\cdot 10^3}{273.375\cdot 10^{-8}}\cdot zc-\frac{3.89\cdot 10^3}{91.125\cdot 10^{-8}}\cdot yc=0\\ &y=\frac{1}{3}z\\ \end{aligned}

Two arbitrary points define a line (neutral axis).

\begin{aligned} &z=0, y=0\\ &z=4.5\sqrt{2}, y=1.5\sqrt{2}\\ \end{aligned}

Stress at suspicious points

\begin{aligned} &\sigma_A=\frac{3.89\cdot 10^3}{91.125\cdot 10^{-8}}\cdot (-4.5\sqrt{2}\cdot 10^{-2})-\frac{3.89\cdot 10^3}{273.375\cdot 10^{-8}}\cdot (1.5\sqrt{2}\cdot 10^{-2})=-181.11\ MPa\\ &\sigma_B=\frac{3.89\cdot 10^3}{91.125\cdot 10^{-8}}\cdot 0-\frac{3.89\cdot 10^3}{273.375\cdot 10^{-8}}\cdot (-3\sqrt{2}\cdot 10^{-2})=181.11\ MPa\\ \end{aligned}

Bending

Strength of materials - Bending - Example 7

Example 7