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Example 1

Design internal force diagrams.

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Solution

Reactions

\begin{aligned} &\sum M_{C . l}=0\\ &-6 \cdot \frac{3}{2}+V_{A} \cdot 3=0\\ &V_{A}:=\frac{6 \cdot \frac{3}{2}}{3}=3\\ &\Sigma X=0\\ &8-H_{F}=0\\ &H_{F}:=8\\ &\Sigma Y=0\\ &V_{A}-6-3 \cdot 4+V_{F}=0\\ &V_{F}:=-\left(V_{A}-6-3 \cdot 4\right)=15\\ &\Sigma M_{A}=0\\ &6 \cdot \frac{3}{2}+8 \cdot 4+3 \cdot 4 \cdot 5+4-V_{F} \cdot 7+M_{F}=0\\ &M_{F}:=-\left(6 \cdot \frac{3}{2}+8 \cdot 4+3 \cdot 4 \cdot 5+4-V_{F} \cdot 7\right)=0 \end{aligned}

Section AB \( 0 \leq x \leq 1.5\)

\begin{aligned} &Q_{A B}=3 \cos (\alpha)=1.8 \quad k N \\ &M_{A B}(x):=3 x \\ &M_{A B}(0)=0 \\ &M_{A B}(1.5)=4.5 \quad k N m \\ &N_{A B}=3 \sin (\alpha)=-2.4 \quad k N \end{aligned}

Section BC \( 1.5 \leq x \leq 3\)

\begin{aligned} &Q_{A B}=3 \cos (\alpha)-6 \cos (\alpha)=-1.8 \mathrm{kN} \\ &M_{A B}(x):=3 x-6 \cdot(x-1.5) \\ &M_{A B}(1.5)=4.5 \quad k N m \\ &M_{A B}(3)=0 \\ &N_{A B}=3 \sin (\alpha)+6 \sin (\alpha)=-2.4 \quad k N \end{aligned}

Graphs

Equilibrium at node C

\begin{aligned} &\Sigma X=0\\ &8-8-2.4 \cdot \cos (\alpha)+1.8 \cdot \sin (\alpha)=0\\ &L=P\\ &\Sigma Y=0\\ &3-2.4 \cdot \sin (\alpha)+1.8 \cdot \cos (\alpha)=0\\ &L=P \end{aligned}