Edupanda » Strength of materials »   Limit Load Capacity   »   Bending   » Example 1

Example 1

Find the limiting value of force P. Given: \(\sigma_P=195 MPa\)

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Solution

\begin{aligned} &\sum M_{A}=0 \\ &6 P \cdot 2-V_{C} \cdot 4-3 P \cdot 7=0 \\ &V_{C}=-\frac{9}{4} P \\ &\sum M_{C}=0 \\ &V_{A} \cdot 4-6 P \cdot 2-3 P \cdot 3=0 \\ &V_{A}=\frac{21}{4} P \end{aligned} \begin{aligned} &A=18 \cdot 2+20 \cdot 1.2+181.6=103.2 \mathrm{~cm}^{2} \\ &\frac{A}{2}=51.6 \mathrm{~cm}^{2} \\ &\frac{A}{2}=18 \cdot 1.6+1.2 x=51.6 \\ &x=19 \\ &S_{y p l 1}=18 \cdot 2.8 \cdot(1+1.4)+1 \cdot 1.2 \cdot 12=121.56 \mathrm{~cm}^{3} \\ &S_{y p l 2}=18 \cdot 1.6 \cdot(19+0.8)+19 \cdot 1.2 \cdot \frac{19}{2}=786.84 \mathrm{~cm}^{3} \\ &W_{p l}=908.4 \mathrm{~cm}^{3} \\ &\left|\frac{M}{W}\right| \leq \operatorname{sigmaP} \\ &10.5 P \leq 908.4 \cdot 10^{-6} \cdot 195 \cdot 10^{6} \\ &P \leq 16.87 \mathrm{kN} \end{aligned}