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Example 1

Freely supported beam with a cross-section shown in the figure below is loaded with a force P=100 kN. Check the stress values at critical points of the dangerous section. Apply the HMH hypothesis in the calculations. Create charts of bending moments and transverse forces, as well as stress distribution along the height of the section. The permissible stresses are: \(k_r=120\ MPa, k_c=80\ MPa\).

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Solution

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\begin{aligned} &z_1=50\ mm &A_1=2000\ mm^2\\ &z_2=110\ mm &A_2=2000\ mm^2\\ &z_c=\frac{z_1\cdot A_1+z_2\cdot A_2}{A_1+A_2} &z_c=80\ mm\\ &I_{yc}=\frac{20\cdot 100^3}{12}+2000\cdot (50 80)^2+\frac{100\cdot 20^3}{12}+2000\cdot (110 80)^2\\ &I_{zc}=5333333\ cm^4=5.33\cdot 10^{6}\ m^4\\ \end{aligned}

Normal stresses

\begin{aligned} &\sigma=\frac{M}{Iy}\cdot z \\ &\sigma=\frac{-7.5\cdot 10^3}{5.33\cdot 10^{-6}}\cdot z=-1.407\cdot 10^9\cdot z\\ &\sigma_1=112.56\ MPa\\ &\sigma_3=0\ MPa\\ &\sigma_2=-28.14\ MPa\\ &\sigma_4=-56.28\ MPa\\ \end{aligned}

Shear stresses

\begin{aligned} &\tau=\frac{T⋅S}{Iy⋅b}\\ &\tau=\frac{75\cdot 10^3}{5.33\cdot 10^{-6}}\cdot \frac{S}{b}=1.407\cdot 10^{10}\cdot \frac{S}{b}\\ \\ \\ \\ &Moment of inertia\\ &S_1=S_4=0\\ &S_3=100⋅20⋅30+20⋅20⋅10=64000\ cm^3=64\cdot 10^{-6}\ m^3\\ &S_2=100⋅20⋅30=60000\ cm^3=60\cdot 10^{-6}\ m^3\\ \\ &\tau_3=1.407⋅10^9⋅\frac{64\cdot 10^{-6}}{0.02}=45.03\ MPa\\ &\tau_2 bis=1.407⋅10^{10}⋅\frac{60\cdot 10^{-6}}{0.02}=42.21\ MPa\\ &\tau_2 prim=8.44\ MPa\\ \end{aligned}

Stress values at critical points of the dangerous section.

\begin{aligned} &\sigma_{red}^{HMH}=\sqrt{\sigma^2+3⋅\tau^2}\\ &\sigma_{red}^{HMH}(2)=\sqrt{28.14^2+3⋅42.21^2}=78.34\ MPa\\ &\sigma_{red}^{HMH}(3)=\sqrt{0^2+3⋅45.03^2}=78\ MPa\\ &\sigma_{red}^{HMH}(1)=\sqrt{112.56^2+3⋅0^2}=112.56\ MPa\\ \end{aligned}

Source:

Stanisław Wolny, Adam Siemieniec, Strength of Materials Part 1, AGH University of Science and Technology Press, Krakow 2002, Example 9.38 p. 382