1. Marking characteristic points and reactions on supports

2. Calculation of reactions using equilibrium equations

\begin{aligned} \\ &\sum{X}=0\\ &H_E+6\cdot4-8=0\\ &H_E=-16kN\\ \\ &\sum{M_A}=0\\ &-6\cdot4\cdot2+10\cdot3-12-6V_E-8H_E=0\\ &V_{E}=16.33kN\\ \\ &\sum{M_{E}}=0\\ &6V_A-8\cdot8+6\cdot4\cdot6-10\cdot3-12=0 \\ &M_{A}=-6.33kN\\ \\ &\sum{Y}=0\\ &V_A+V_E-10=0\\ &L=P\\ \end{aligned}

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

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

\begin{aligned} &Q_{AB}=-8+6x\\ &Q_{A}(0)=-8kN\\ &Q_{B}(4)=16kN\\ &M_{AB}=-8x+6x\frac{x}{2}\\ &M_{A}(0)=0\\ &M_{B}(4)=16kNm\\ &N_{AB}=-6.33kN\\ \end{aligned}

extremum

\begin{aligned} &Q_{AB}=-8+6x=0\\ &8=6x\\ &x=1.33m\\ &M_{max}=-5.33kNm\\ \end{aligned}

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

\begin{aligned} &Q_{BC}=-6.33kN\\ &M_{BC}=-6.33x-8\cdot4+24\cdot2\\ &M_{B}(0)=16kNm\\ &M_{C}(3)=-3kNm\\ &N_{BC}=8-24=-16kN\\ \end{aligned}

c) Interval ED \(x \in{\langle 0,4)}\)

\begin{aligned} &Q_{ED}=16kN\\ &M_{ED}=-16x\\ &M_{E}(0)=0\\ &M_{D}(4)=-64kNm\\ &N_{ED}=-16.33kN\\ \end{aligned}

d) Interval DC \(x \in{\langle 0,3)}\)

\begin{aligned} &Q_{DC}=-16.33kN\\ &M_{DC}=-16\cdot4+16.33x+12\\ &M_{D}(0)=-52kNm\\ &M_{C}(3)=-3kNm\\ &N_{DC}=-16kN\\ \end{aligned}

4. Final diagrams

Internal Forces - Frames

Strength of materials - Internal Forces - Frames - Example 12

Example 12