1. Marking characteristic points and reactions on supports

2. Calculating reactions using equilibrium equations

\begin{aligned} \\ &\sum{M_{D}}=0\\ &-20 + 5V_{B}- 6\cdot 2 - 6\cdot 2=0\\ &V_{B}=8.8 \ kN\\ \\ &\sum{M_{B}}=0\\ &-20 - 5V_{D} + 6\cdot 3 + 8\cdot 4\cdot 5 - 6\cdot 7 = 0\\ &V_{D}=23.2 \ kN\\ \\ &\sum{Y}=0\\ &V_{B} + V_{D} - 6 + 6 - 8\cdot 4=0\\ &L=P\\ \\ &\frac{2.8}{x}=\frac{13.2}{2-x} \ \ \ \ \Rightarrow \ \ \ \ x=0.35\\ &\frac{10}{x}=\frac{6}{2-x} \ \ \ \ \ \Rightarrow \ \ \ \ \ \ x=1.25\\ \\ \end{aligned}

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

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

\begin{aligned} &Q_{AB}=0\\ &M_{AB}=-20 \ kNm\\ \end{aligned}

b) Interval BC \(x \in{\langle 3,6)}\)

\begin{aligned} &Q_{BC}=8.8 \ kN\\ &M_{BC}=-20 +8.8(x_{2}-3)\\ &M_{B}(3)=-20 \ kNm\\ &M_{C}(6)=6.44 \ kNm\\ \end{aligned}

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

\begin{aligned} &Q_{ED}=-6+8x_{3}\\ &Q_{E}(0)=-6 \ kN\\ &Q_{D}(2)=10 \ kN\\ &Q_{ED}=0\\ &-6 + 8x_{3}=0\\ &x_{3}=\frac{6}{8}=0.75\\ &M_{ED}= 6x-\frac{1}{2}\cdot 8x_{3}^{2}\\ &M_{E}(0)=0\\ &M_{D}(2)=-4 \ kNm\\ &M_{max}(0.75)=2.25 \ kNm\\ \end{aligned}

d) Interval DC \(x \in{\langle 2,4)}\)

\begin{aligned} &Q_{DC}=-6 + 8x_{4} - 23.2\cdot 8\\ &Q_{D}(2)=-13.2 \ kN\\ &Q_{C}(4)=2.8 \ kN\\ &Q_{DC}=-29.2 + 8x = 0\\ &x=3.65\\ &M_{DC}= 6x_{4}= \frac{1}{2}\cdot 8 \cdot x_{4} + 23.2(x_{4}-2)\\ &M_{D}(2)=-4.4 \ kNm\\ &M_{C}(4)=6.44 \ kNm\\ &M_{max}(3.65)=6.89 \ kNm\\ \end{aligned}

4. Final graphs

Internal Forces - Beams

Strength of materials - Internal Forces - Beams - Example 3

Example 3