Content

Determine and draw the internal force diagrams.

Video solution

Solution

1. Identification of reactions at supports

2. Calculation of reactions using equilibrium equations

\begin{aligned} &\sum{X}=0\\ &H_A+10=0\\ &H_A=-10 \text{ kN}\\ \\ &\sum{M_{A}}=0\\ &10 \cdot 3\cdot 1,5 +10 \cdot 3 -V_C \cdot 3=0\\ &V_{C}=25 \text{ kN}\\ \\ &\sum{Y}=0\\ &V_A-10\cdot 3+V_C=0 \\ &V_{A}=5 \text{ kN}\\ \end{aligned}

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

a) Interval AB \(x \in{\langle0,3)}\)

\( \quad \quad \)

\begin{aligned} &Q_{AB}=10 \text{ kN}\\ &M_{AB}=10x\\ &M_{A}(0)=0 \text{ kNm}\\ &M_{B}(3)=30 \text{ kNm}\\ &N_{AB}=-5 \text{ kN}\\ \end{aligned}

b) Interval CB \(x \in{\langle0,3)}\)

\( \quad \quad \)

\begin{aligned} &Q_{CB}=-25+10\cdot x\\ &Q_C(0)=-25 \text{ kN}\\ &Q_B(3)=5 \text{ kN}\\ &M_{CB}=25\cdot x-10\cdot x\cdot \frac{x}{2}\\ &M_C(0)=0 \text{ kNm}\\ &M_B(3)=30 \text{ kNm}\\ &N_{CB}=10 \text{ kN}\\ \end{aligned}

4. Final diagrams

Internal forces - frames

Strength of materials - Internal forces - frames - Example 1

Example 1