Reactions

\begin{aligned} &\sum{M_{A}}=0 & 6\cdot 6\cdot 3-6\cdot V_B+10\cdot 8=0 & V_B=31,33\ kN\\ &\sum{M_{B}}=0 & V_A\cdot 6-6\cdot 6\cdot 3+10\cdot 2=0 & V_A=14,67\ kN\\ &\sum{Y}=0 & V_A+V_B-6\cdot 6-10=0 & \ L=P\\ \end{aligned}

Diagram of shear forces and bending moments

Dimensioning

We know that for an I-beam, the distance from the center of gravity to the upper fibers will be LESS than the distance to the lower fibers, which means that greater stresses will occur on the lower fibers. In the alpha-alpha section, the bending moment is less than the bending moment in the beta-beta section. In the alpha-alpha section, the lower fibers are in tension, meaning they are inherently more loaded. The tensile strength is less than the compressive strength, so this condition may be decisive, because not only are the tensile stresses the maximum stresses in this section, but also the tensile strength is less than the compressive strength. Although the situation is reversed in the beta-beta section, this section also needs to be checked because the bending moment is greater.

\begin{aligned}\\ &z_1=6,5a & A_1=5a^2\\ &z_2=3a & A_2=6a^2\\ &z_c=\frac{z_1\cdot A_1+z_2\cdot A_2}{A_1+A_2}=4,59a\\ &I_{y_c}=I_{y_1}+A_1\cdot (z_1-z_c)^2+I_{y_2}+A_2\cdot (z_2-z_c)^2=\\ &\frac{5a\cdot (1a)^3}{12}+5a^2\cdot (6,5a-4,59a)^2+\frac{1a\cdot (6a)^3}{12}+6a^2\cdot (3a-4,59a)^2=51,826a^4\\ &W_A=\frac{I_{y_c}}{z_A}=\frac{51,826}{4,59}a^3=11,29a^3\\ &W_B=\frac{I_{y_c}}{z_B}=\frac{51,826}{7-4,59}a^3=21,50a^3\\ \end{aligned}

The stresses in the tensile fibers (A) are compared to the tensile strength, and in the compressed fibers (B) – to the compressive strength.

\begin{aligned}\\ &\alpha – \alpha \\ &\frac{M_{\alpha \alpha}}{W_A}\le k_r & \frac{18\cdot 10^3}{11,29a^3}\le 120\cdot 10^6 & a\ge 0,0237\\ &\frac{M_{\alpha \alpha}}{W_B}\le k_c & \frac{18\cdot 10^3}{21,5a^3}\le 160\cdot 10^6 & a\ge 0,0174\\ \end{aligned}

Here the situation is reversed, the upper fibers (B) are in tension, and the lower fibers (A) are in compression.

\begin{aligned}\\ &\beta – \beta\\ &|\frac{M_{\beta \beta}}{W_A}|\le k_c & \frac{20\cdot 10^3}{11,29a^3}\le 160\cdot 10^6 & a\ge 0,0223\\ &|\frac{M_{\beta \beta}}{W_B}|\le k_r & \frac{20\cdot 10^3}{21,5a^3}\le 120\cdot 10^6 & a\ge 0,0198\\ \\ &a=0,025\ m = 25\ mm\\ \end{aligned}

For the dimensioned section, the stresses are:

\begin{aligned}\\ &\alpha – \alpha \\ &\sigma_A=102,04\ MPa\\ &\sigma_B=53,58\ MPa\\ \\ &\beta – \beta \\ &\sigma_A=113,375\ MPa\\ &\sigma_B=59,53\ MPa\\ \\ \end{aligned}

Stress diagrams

Bending

Strength of materials - Bending - Example 9

Example 9