Ejemplo 1

Design and draw internal force diagrams.

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Solución

1. Designation of characteristic points and reactions at supports

2. Calculation of reactions using equilibrium equations

\begin{aligned} &\sum X=0 \\ &-6+H_{D}=0 \\ &H_{D}=6 k N \\ &\sum Y=0 \\ &2-2 \cdot 6+6+V_{B}=0 \\ &B_{B}=4 k N \\ &\sum M_{B}=0 \\ &2 \cdot 3+6-6 \cdot 4+6 \cdot 2 \cdot 3-M_{D}=0 \\ &M_{D}=24 k N \end{aligned}

3. Writing equations for internal forces in individual variable intervals:

a) Interval AB x \in{\langle 0,3)}

\begin{aligned} &Q_{AB}=2 \ kN\\ &M_{AB}=2x\\ &M_{A}(0)=0\\ &M_{B}(3)=6 \ kNm\\ \end{aligned}

b) Interval BC x \in{\langle 3,7)}

\begin{aligned} &Q_{BC}=2 + 4 – 2(x-3)\\ &M_{BC}=2x + 4(x-3) + 6 – \frac{1}{2}\cdot 2(x-3)^{2}\\ &Q_{B}(3)=6 \ kN\\ &Q_{C}(7)=2 \ kN\\ &Q_{BC}=0\\ &6-2(x-3)=0\\ &6-2x + 6=0\\ &x=6\\ &M_{B}(3)=12 \ kNm\\ &M_{max}(6)=21 \ kNm\\ &M_{C}(7)=20 \ kNm\\ \end{aligned}

c) Interval DC x \in{\langle 0,2)}

\begin{aligned} &Q_{DC}=2x\\ &Q_{D}(0)=0\\ &Q_{C}(2)=4 \ kN\\ &M_{DC}=24 – 2x\frac{x}{2}=24 – x^{2}\\ &M_{D}(0)=24 \ kNm\\ &M_{C}(2)=20 \ kNm\\ &N_{DC}=6 \ kN\\ \end{aligned}

4. Final diagrams