Content

Determine and draw the internal force diagrams.

Solution

1. Marking characteristic points and reactions at supports

2. Calculating reactions using equilibrium equations

\begin{array}{ll} \Sigma X=0 & -H_A+10=0 & H_A=10 \text{kN} \\ \Sigma M_A=0 & 10+5 \cdot 4 \cdot 2+10 \cdot 5+10 \cdot 4-6 V_E=0 & V_E=23,33 \ \mathrm{kN} \\ \Sigma M_E=0 & -10 \cdot 2+10 \cdot 2-5 \cdot 4 \cdot 4+10+6 V_A+3 H_A=0 & V_A=6,67 \ \mathrm{kN} \\ \Sigma Y=0 & V_A+V_E-5 \cdot 4-10=0 & L=P \\ \end{array}

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

a) Interval AB (0 < x < 5)

\( \quad \quad \)

\begin{aligned} & Q_{A B}=10 \ kN\\ & M_{A B}=10 x \\ & M(0)=0 \ \mathrm{kNm} \\ & M(5)=50 \ \mathrm{kNm} \\ & N_{A B}=-6,67 \ \text{kNm} \\ \end{aligned}

b) Interval BC (0 < x < 4)

\( \quad \quad \)

\begin{aligned} & Q_{B C}=6,67-5 x \\ & Q(0) \rightarrow 6,67 \ kN \\ & Q(4) \rightarrow-13,33 \ kN\\ & M_{B C}=10 \cdot 5+6,67 x+10-5 \cdot x \frac{x}{2} \\ & M(0) \rightarrow 60 \ kNm \\ & M(4) \rightarrow 46,68 \ kNm \\ & N_{B C}=10 \ kN\\ \\ & \text{Ekstremum:} \\ & Q(x)=0 \\ & 5 x=6,67 \quad x=1,334 \ m \\ & M_{max}(1,334)=64,45 \ kNm \\ \end{aligned}

c) Interval ED (0 < x < 2)

\( \quad \quad \)

\begin{aligned} & Q_{ED}=-23,33 \ kN\\ & M_{E D}=23,33 \cdot x \\ & M(0) \rightarrow 0 \ kNm \\ & M(2) \rightarrow 46,66 \ \mathrm{kNm} \\ & N_{ED}=0 \\ \end{aligned}

d) Interval DC (0 < x < 2)

\( \quad \quad \)

\begin{aligned} & Q_{DC}=0 \ kN\\ & M_{E D}=46,66 \ kNm \\ & N_{ED}=-13,33 \ kN \\ \end{aligned}

4. Final diagrams

Internal forces - frames

Strength of materials - Internal forces - frames - Example 5

Example 5