PROCEDURE FLOW - method of forces

SSN=4-3=1

Basic arrangement of the method of forces (UPMS)

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} \\ &x_{3} \in(0 ; 4 l) \\ &M_{1}\left(x_{3}\right)=3 l \end{aligned}

State P

Bending moment function

\begin{aligned} &x_{1} \in(0 ; 2 l) \\ &M_{p}\left(x_{1}\right)=-P x_{1} \\ &x_{2} \in(0 ; 3 l) \\ &M_{p}\left(x_{2}\right)=-P\left(x_{2}+2 l\right) \\ &x_{3} \in(0 ; 4 l) \\ &M_{p}\left(x_{3}\right)=-5 P l \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+\int_{0}^{4 l} 9 l^{2} d x=\frac{1}{3} 27 l^{3}+36 l^{3}=45 l^{3} \\ \delta_{11}=\frac{45 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}(-P(x+2 l) \cdot x) d x+\int_{0}^{4 l}\left(-15 P l^{2}\right) d x=\int_{0}^{3 l}\left(-P x^{2}-2 P l x\right) d x-60 P l^{3}= \\ \delta_{1 p}=-\frac{78 P l^{3}}{E I}(3 l)^{3}-2 P l \cdot \frac{1}{2}(3 l)^{2}-60 P l^{3}=-78 P l^{3} \\ \\ =-P \cdot \frac{1}{3} \end{gathered}

After calculating the statically indeterminate reactions, we calculate and plot the final charts.

\begin{aligned} &x_{1} \in(0 ; 2 l) \\ &N\left(x_{1}\right)=0 \\ &T\left(x_{1}\right)=P \\ &M\left(x_{1}\right)=-P x \\ &M(0)=0 \\ &x_{2} \in(0 ; 3 l) \\ &N\left(x_{2}\right)=0 \\ &T\left(x_{2}\right)=P-\frac{26}{15} P=-\frac{11}{15} P \\ &M\left(x_{2}\right)=\frac{26}{15} P x-P(x+2 l) \\ &M(0)=-2 P l \\ &x_{3} \in(0 ; 4 l) \\ &N\left(x_{3}\right)=-\frac{26}{15} P+P=-\frac{11}{15} P \\ &T\left(x_{3}\right)=0 \\ &M\left(x_{3}\right)=0,2 P l \end{aligned}

Final internal force charts

Energy Methods - Statically Indeterminate Systems

Strength of materials - Energy Methods - Statically Indeterminate Systems - Example 2

Example 2