1. Identifying key points and reactions at supports

2. Calculating reactions using equilibrium equations

\begin{aligned} \\ &\sum{X}=0\\ &10-10\cdot3+H_E=0\\ &H_E=20kN\\ \\ &\sum{Y}=0\\ &V_A-12\cdot2+12=0\\ &V_{A}=12kN\\ \\ &\sum{M_{A}}=0\\ &M_A-10\cdot3\cdot\frac{3}{2}+10+12\cdot2\cdot4-12\cdot5+H_E\cdot1=0 \\ &M_{A}=-21kNm\\ \\ &\sum{M_E}=0\\ &-12\cdot2\cdot1+10-10\cdot3\cdot0,5-10\cdot1+5V_A+M_A=0\\ &L=P\\ \end{aligned}

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

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

\begin{aligned} &Q_{AB}=-10+10x\\ &Q_{A}(0)=-10kN\\ &Q_{B}(3)=20kN\\ &M_{AB}=-10x+10x\frac{x}{2}-21\\ &M_{A}(0)=-21kNm\\ &M_{B}(3)=-6kNm\\ &N_{AB}=-12 kN\\ \end{aligned}

extremum

\begin{aligned} &Q_{AB}=-10+10x=0\\ &10=10x\\ &x=1m\\ &M_{max}=-26kNm\\ \end{aligned}

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

\begin{aligned} &Q_{BC}=12kN\\ &M_{BC}=-10\cdot3+30\cdot\frac{3}{2}-21+12x\\ &M_{B}(0)=-6kNm\\ &M_{C}(3)=30kNm\\ &N_{BC}=20kN\\ \end{aligned}

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

\begin{aligned} &Q_{ED}=-12+12x\\ &Q_{E}(0)=-12kN\\ &Q_{D}(2)=12kN\\ &M_{ED}=12x-\frac{1}{2}12x^2\\ &M_{E}(0)=0\\ &M_{D}(2)=0\\ &N_{ED}=20kN\\ \end{aligned}

extremum

\begin{aligned} &Q_{ED}=-12+12x=0\\ &12=12x\\ &x=1m\\ &M_{max}(1)=6kNm\\ \end{aligned}

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

\begin{aligned} &Q_{DC}=-20kN\\ &M_{DC}=12\cdot2-12\cdot2\cdot1+20x\\ &M_{D}(0)=0\\ &M_{C}(2)=40kNm\\ &N_{DC}=12kN\\ \end{aligned}

4. Final graphs

Fuerzas internas - Marcos

Resistencia de materiales - Fuerzas internas - Marcos - Ejemplo 8

Ejemplo 8