Example 2

Design and draw internal force diagrams.

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Solution

Classic version

1. Marking characteristic points and reactions at supports

\begin{aligned} &\sum M_{D}=0 \\ &10 \cdot 8-60 \cdot 5-15+4 V_{B}=0 \\ &V_{B}=58.75 \mathrm{kN} \\ &\sum M_{B}=0 \\ &10 \cdot 4-60 \cdot 1-15-4 V_{D}=0 \\ &V_{D}=-8.75 \mathrm{kN} \\ &\sum Y=0 \\ &V_{B}+V_{D}-60+10=0 \\ &L=P \end{aligned}

3. Decomposition of internal force equations in individual variability intervals:

a)Interval AB x \in{\langle 0,4)}

\begin{aligned} &Q_{A B}=10-10 \cdot x \\ &Q_{A(0)}=10 \\ &Q_{B(4)}=-30 \\ &M_{A B}=10 \cdot x-10 \cdot \frac{x^{2}}{2} \\ &M_{A(0)}=0 \\ &M_{B(4)}=-40 \end{aligned}

\begin{aligned} \\ &Q_{AB}=10-10\cdot x=0\\ &10=10x\\ &x=1m\\ \\ \end{aligned}

Extremum

\begin{aligned} \\ &\frac{10}{x}=\frac{30}{4-x} \ \Rightarrow \ x=1\\ &M_{max}=10\cdot 1 – 10\cdot(\frac{1}{2})^{2}=5\\ \\ \end{aligned}

b) Interval BC x \in{\langle 4,6)}

\begin{aligned} \\ &Q_{BC}=10+V_{B}-10\cdot x\\ &Q_{B(4)}=28,75\\ &Q_{C(6)}=8,75\\ &M_{BC}=10\cdot x + V_{B}(4-x)\\ &M_{B(4)}=-40\\ &M_{C(6)}=-2,5\\ \\ \end{aligned}

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

\begin{aligned} \\ &Q_{DC}=-V_{D}=8,75\\ &M_{DC}=V_{D}\cdot x\\ &M_{D(0)}=0\\ &M_{C(2)}=17,5\\ \\ \end{aligned}

4. Final graphs