Example 1

For a given arrangement, determine the internal force diagrams using the displacement method.

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Solution

1. Determination of the degree of kinematic indeterminacy:

\begin{aligned} &SKN=1\\ \end{aligned}

2. Selection of the basic system of the displacement method (UPMP):

3. Graphs and determination of the coefficients and free terms of the equation:

Determination of the coefficients of UPMP

\begin{aligned} &r_{11}=\frac{2EI}{3} +\frac{3EI}{4} =\frac{17EI}{12}\\ &r_{1p}=12-9=3\\ \end{aligned}

System of equations:

\begin{aligned} &r_{11}\cdot\varphi_{1} +r_{1p}=0\\ \end{aligned} \begin{aligned} &\varphi_{1}=-\frac{r_{1p}}{r_{11}}=-\frac{36}{17EI}\\ \\ &M_{ost}=M_{p}+\varphi_{1}\cdot M_{1}\\ &M_{AB}=-12-\frac{1}{3}\cdot \left(-\frac{36}{17}\right)=-12,706\\ &M_{BA}=12+\frac{2}{3}\cdot \left(-\frac{36}{17}\right)=-10,588\\ &M_{BC}=-9+\frac{3}{4}\cdot \left(-\frac{36}{17}\right)=-10,588\\ &M_{CB}=0\\ \end{aligned}

Element AB:

\begin{aligned} &\sum{M_{B}}=0\\ &{Q_{AB}}\cdot 6+10,588-12,7-4\cdot6\cdot3=0\\ &{Q_{AB}}=12,352\\ &\sum{M_{A}}=0\\ &{Q_{BA}}\cdot 6+10,588+4\cdot 6\cdot 3-12,7=0\\ &{Q_{BA}}=-11,648\\ &\frac{12,352}{x}=\frac{11,648}{6-x}\\ &74,112-12,352x=11,648\\ &x=3,088\\ &M_{max}=-12,7+0,5\cdot 12,352\cdot 3,088\\ \end{aligned}

Element BD:

\begin{aligned} &\sum{M_{C}}=0\\ &{Q_{BC}}\cdot 4-12\cdot 2-10,588=0\\ &{Q_{BC}}=8,647\\ &\sum y=0\\ &8,647-12-Q_{CD}=0\\ &Q_{CD}=-3,353\\ \end{aligned}

Final graphs of internal forces