1. Marking characteristic points and reactions on supports

2. Calculation of reactions using equilibrium equations

\begin{aligned} &\sum{M_{A}}=0\\ &20 - 12\cdot 2 - V_{C}\cdot 4 + 8\cdot 6\cdot 5=0\\ &V_{C} = 59 \ kN\\ &\sum{M_{C}}=0\\ &20 + V_{A} \cdot 4 + 12\cdot 2 + 8\cdot 6\cdot 1=0\\ &V_{A}=-23 \ kN\\ &\sum{Y}=0\\ &V_{A}+V_{C} - 9\cdot 6 + 12 = 0\\ &L=P\\ \end{aligned}

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

a) Range AB \( x \in{\langle 0,2)} \)

\begin{aligned} &Q_{AB}=-23 \ kN\\ &M_{AB}=20 - 23x\\ &M_{A}(0)=20 \ kNm\\ &M_{B}(0)=-26 \ kNm\\ \end{aligned}

b) Range BC \(x \in{\langle 2,4)}\)

\begin{aligned} &Q_{BC}=-23 + 12 - 8(x-2)\\ &Q_{B}(2)=-11 \ kN\\ &Q_{C}(4)=-27 \ kN\\ &M_{BC}=-23x + 20 + 12(x_{2}-2) - \frac{1}{2}\cdot 8(x-2)^{2}\\ &M_{B}(2)=-26 \ kNm\\ &M_{C}(4)=-64 \ kNm\\ \end{aligned}

c) Range DC \(x \in{\langle 0,4)}\)

\begin{aligned} &Q_{DC}=8x\\ &Q_{D}(0)=0\\ &Q_{C}(4)=32 \ kN\\ &M_{DC}=-\frac{1}{2}8x^{2}\\ &M_{D}(0)=0\\ &M_{C}(4)=-64 \ kNm\\ \end{aligned}

4. Final graphs

Internal Forces - Beams

Strength of materials - Internal Forces - Beams - Example 4

Example 4