Example 1

Determine and draw internal force diagrams.

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Solution

Classic Version

1. Marking characteristic points and reactions on supports

2. Calculation of reactions using equilibrium equations

\begin{aligned} &\sum{X}=0\\ &H_A=0\\ \end{aligned} \begin{aligned} &\sum{M_{A}}=0\\ &10\cdot 2+15\cdot 5-V_{D}\cdot 8=0\\ &V_{D}=11,875 \ kN\\ \end{aligned} \begin{aligned} &\sum{M_{D}}=0\\ &V_{A}\cdot 8-15\cdot 3-10\cdot 6=0\\ &V_{A}=13,125 \ kN\\ \end{aligned} \begin{aligned} &\sum{Y}=0\\ &V_{A}+V_{D}-10-15=0\\ &L=P\\ \end{aligned}

3. Expanding the internal force equations in individual sections of variability:

a) Section AB

\begin{aligned} &Q_{AB}=V_{A}=13.125 kN\\ &M_{AB}=V_{A}\cdot x\\ &M_{A}(0)=0\\ &M_{B}(2)=26.25 kNm\\ \end{aligned}

b) Section BC x \in{\langle 2,5)}

\begin{aligned} &Q_{BC}=V_{A}-10=3.125\\ &M_{BC}=V_{A}\cdot x-10\cdot (x-2)\\ &M_{B}(2)=26.25 kNm\\ &M_{C}(5)=35.625 kNm\\ \end{aligned}

c) Section DC x \in{\langle 0,3)}

\begin{aligned} &Q_{DC}=-V_{D}\\ &M_{DC}=V_{D}\cdot x\\ &M_{D(0)}=0\\ &M_{C(3)}=35.625 kNm\\ \end{aligned}

4. Final plots