The moment vector is perpendicular to the plane of the load (drawing below).

The value of the moment at the most loaded section is 2 kNm.

This moment is decomposed into principal central directions.

In the previous example, we immediately marked the principal central axes at a 45-degree angle. You can also calculate the central principal moments first, the deviation moment, and the rotation angle to the principal central axes as shown below.

\begin{aligned} &I_y=I_z=\frac{9^4}{36}=182.25\ cm^4\\ &D=\frac{9^4}{72}=91.125\ cm^4\\ &I_{1,2}=\frac{I_y+I_z}{2}\pm \sqrt{(\frac{I_y-I_z}{2})^2+D^2}\\ &I_1=273.375\ cm^4\\ &I_2=91.125\ cm^4\\ &tg_{\varphi_1}=\frac{D}{I_z-I_1}\Rightarrow \varphi_1=-45^o\\ \\ \\ &M_1=M_2=M=M_y\cdot \frac{\sqrt{2}}{2}=\sqrt{2}\ kNm\\ &\sigma=\frac{M_1}{I_1}z-\frac{M_2}{I_2}y=M\cdot (\frac{z}{I_1}-\frac{y}{I_2})\\ &neutral\ axis\ \sigma=0\\ &M\cdot (\frac{z}{I_1}-\frac{y}{I_2})=0\\ &z=\frac{y\cdot I_1}{I_2}\\ &z=3y\\ \end{aligned}

Two arbitrary points define a straight line (neutral axis).

\begin{aligned} &y=0, z=0\\ &y=1, z=3\\ \end{aligned}

Normal stresses

Coordinates of points in the principal central axes system

\begin{aligned} &\sigma=\frac{\sqrt{2}\cdot 10^3}{273.375\cdot 10^{-8}}\cdot z\cdot 10^{-2}-\frac{\sqrt{2}\cdot 10^3}{91.125\cdot 10^{-8}}\cdot y\cdot 10^{-2}\\ &\sigma=5.173\cdot 10^6\cdot z-15.519\cdot 10^6\cdot y=5.173z-15.519y\ [MPa]\\ \\ &A (-1.5\sqrt{2}; 4.5\sqrt{2})\\ &\sigma_A=5.173\cdot 4.5\sqrt{2}-15.519\cdot (-1.5\sqrt{2})=65.84\ MPa\\ \\ &B (3\sqrt{2}; 0)\\ &\sigma_B=-15.519\cdot (3\sqrt{2})=-65.84\ MPa\\ \end{aligned}

Bending

Strength of materials - Bending - Example 8

Example 8