PROCEDURE – method of forces

SSN=4-3=1

Basic arrangement of the method of forces (BAMF)

State x1=1

Bending moment function

\begin{aligned} &x_{1} \in(0 ; 2 l) \\ &M_{1}\left(x_{1}\right)=0 \\ &x_{2} \in(0 ; 3 l) \\ &M_{1}\left(x_{2}\right)=1 \cdot x_{2} \end{aligned}

Bending moment function

\begin{aligned} &x_{1} \in(0 ; 2 l) \\ &M_{p}\left(x_{1}\right)=-\frac{1}{2} q x_{1}^{2} \\ &x_{2} \in(0 ; 3 l) \\ &M_{p}\left(x_{2}\right)=-2 q l\left(x_{2}+l\right)-\frac{1}{2} q x_{2}^{2} \end{aligned}

Canonical equation of the method of forces

\begin{gathered} \delta_{11} \cdot x_{1}+\delta_{1 p}=0 \\ E I \delta_{11}=\int_{0}^{L}\left(M_{1} \cdot M_{1}\right) d x=\int_{0}^{3 l} x^{2} d x=\frac{1}{3} 27 l^{3}=9 l^{3} \\ \delta_{11}=\frac{9 l^{3}}{E I} \\ E I \delta_{1 p}=\int_{0}^{L}\left(M_{1} \cdot M_{p}\right) d x=\int_{0}^{3 l}\left(-2 q l x(x+l)-\frac{1}{2} q x^{3}\right) d x=\int_{0}^{3 l}\left(-2 q l x^{2}-2 q l^{2} x-\frac{1}{2} q x^{3}\right) d x= \\ -2 q l \cdot \frac{1}{3} \cdot 27 l^{3}-2 q l^{2} \cdot \frac{1}{2} \cdot 9 l^{2}-\frac{1}{2} q \cdot \frac{1}{4} \cdot 81 l^{4}=-\frac{297}{8} q l^{4} \\ \delta_{1 p}=-\frac{297}{8 E I} q l^{4} \\ x_{1}=\frac{\frac{297}{8} q l^{4}}{9 l^{3}}=\frac{33}{8} q l \end{gathered}

After calculating the statically indeterminate reaction, we calculate the remaining reactions using the static equilibrium equations.

\begin{aligned} &\sum y=0 \Rightarrow V_{A}=\frac{7}{8} q l \\ &\sum x=0 \Rightarrow H_{A}=0 \\ &\sum M_{A} \Rightarrow M_{A}=-\frac{1}{8} q l^{2} \end{aligned}

Energy Methods - Statically Indeterminate Systems

Strength of materials - Energy Methods - Statically Indeterminate Systems - Example 4

Example 4