1. Marking characteristic points and reactions on supports

2. Calculation of reactions using equations of equilibrium

\begin{aligned} \\ &\sum{X}=0\\ &H_A-2-3=0\\ &H_E=5kN\\ \\ &\sum{Y}=0\\ &V_A-6+3=0\\ &V_{A}=3kN\\ \\ &\sum{M_{A}}=0\\ &M_A-5+3\cdot2\cdot1-1\cdot2\cdot4-3\cdot4-3\cdot3=0 \\ &M_{A}=28kNm\\ \\ &\sum{M_C}=0\\ &M_A-5-3\cdot2\cdot1+V_A\cdot2-H_A\cdot5+1\cdot2\cdot1+3\cdot2-3\cdot2=0\\ &L=P\\ \end{aligned}

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

a) Interval AB \(x \in{\langle 0,5)}\)

\begin{aligned} &Q_{AB}=-5kN\\ &M_{AB}=28-5x\\ &M_{A}(0)=28kNm\\ &M_{B}(5)=3kNm\\ &N_{AB}=-3 kN\\ \end{aligned}
b) Interval BC \)x \in{\langle 0,2)}\)

\begin{aligned} &Q_{BC}=3-3x\\ &Q_{B}(0)=3kN\\ &Q_{C}(2)=-3kN\\ &M_{BC}=28-5+3x-5\cdot5-3x\frac{x}{2}\\ &M_{B}(0)=-2kNm\\ &M_{C}(2)=-2kNm\\ &N_{BC}=-5kN\\ \end{aligned}

extrema

\begin{aligned} &Q_{BC}=3-3x=0\\ &3=3x\\ &x=1m\\ &M_{max}(1)=-0.5kNm\\ \end{aligned}

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

\begin{aligned} &Q_{ED}=-3kN\\ &M_{ED}=3x\\ &M_{E}(0)=0\\ &M_{D}(2)=6kNm\\ &N_{ED}=-3kN\\ \end{aligned}

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

\begin{aligned} &Q_{DC}=3+x\\ &Q_{D}(0)=3kN\\ &Q_{C}(2)=5kN\\ &M_{DC}=-3x+6-x\frac{x}{2}\\ &M_{D}(0)=6kNm\\ &M_{C}(2)=-2kNm\\ &N_{DC}=-3kN\\ \end{aligned}

4. Final charts

Internal Forces - Frames

Strength of materials - Internal Forces - Frames - Example 9

Example 9