Solution

Content

Calculate and draw internal force diagrams.

Solution

1. Marking reactions on supports

2. Calculation of reactions using equilibrium equations

\begin{aligned} \\ &\sum{X}=0\\ &H_A+10=0\\ &H_A=-10kN\\ \\ &\sum{M_{A}}=0\\ &10 \cdot 3\cdot 1.5 +10 \cdot 3 -V_C \cdot 3=0\\ &V_{C}=25kN\\ \\ &\sum{Y}=0\\ &V_A-10\cdot 3+V_C=0 \\ &V_{A}=5kN\\ \end{aligned}

3. Writing equations for internal forces in different intervals of variation:

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

\begin{aligned} &Q_{AB}=10 kN\\ &M_{AB}=10x\\ &M_{A}(0)=0\\ &M_{B}(3)=30 kNm\\ &N_{AB}=-5 kN\\ \end{aligned}

b) Interval CB x \in{\langle 0,3)}

\begin{aligned} &Q_{CB}=-25+10\cdot x\\ &Q_C(0)=-25\ kN\\ &Q_B(3)=5\ kN\\ &M_{CB}=25\cdot x-10\cdot x\cdot \frac{x}{2}\\ &M_C(0)=0\ kNm\\ &M_B(3)=30\ kNm\\ &N_{CB}=10\ kN\\ \end{aligned}

4. Final diagrams