SSN=4-3=1
Basic arrangement of the force method (UPMS)

State x1=1

Bending moment function

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

State P

Bending moment function

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

Canonical equation of the force method

\begin{aligned} &\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}^{l} x_{1}^{2} d x+\int_{0}^{2 l}\left(x_{2}+l\right)^{2} d x=\\ &\frac{1}{3} l^{3}+\int_{0}^{2 l}\left(x^{2}+2 x l+l^{2}\right) d x=\frac{1}{3} l^{3}+\frac{1}{3} \cdot 8 l^{3}+2 l \cdot \frac{1}{2} \cdot 4 l^{2}+l^{2} \cdot 2 l=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}^{2 l}\left(-\frac{1}{2} q x^{2} \cdot(x+l)\right) d x=\int_{0}^{2 l}\left(-\frac{1}{2} q x^{3}-\frac{1}{2} q l x^{2}\right) d x=-\frac{1}{2} q \cdot \frac{1}{4} \cdot 16 l^{4}-\frac{1}{2} q l \cdot \frac{1}{3} \cdot 8 l^{3}=-\frac{10}{3} q l^{4}\\ &\delta_{1 p}=-\frac{10 q l^{4}}{3 E I}\\ &x_{1}=\frac{\frac{10}{3} q l^{4}}{9 l^{3}}=\frac{10}{27} q l \end{aligned}

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

\begin{aligned} &\sum y=0 \Rightarrow V_{A}=\frac{44}{27} q l \\ &\sum x=0 \Rightarrow H_{A}=0 \\ &\sum M_{A} \Rightarrow M_{A}=-\frac{8}{9} q l^{2} \end{aligned}

Energy Methods - Statically Indeterminate Systems

Strength of materials - Energy Methods - Statically Indeterminate Systems - Example 6

Example 6