1. Determination of the degree of kinematic indeterminacy:

\begin{aligned} &SKN=\sum \varphi +\sum \Delta\\ &SKN=1+1=2\\ \end{aligned}

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

Canonical equations of the displacement method:
\( r_{11}\cdot\varphi_{1} +r_{12}\cdot\Delta_{2}+r_{1p}=0\\ r_{21}\cdot\varphi_{1} +r_{22}\cdot\Delta_{2}+r_{2p}=0\\ \)

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

Determination of UPMP coefficients

\begin{aligned} &r_{11}=\frac{3EI}{5}+\frac{3EI}{4}=\frac{27EI}{20}\\ \\ \\ &r_{21}=-\frac{3EI}{16}\\ \end{aligned}

Determination of UPMP coefficients

\begin{aligned} &r_{12}=-\frac{3EI}{16}\\ \\ \\ &r_{22}=\frac{3EI}{32}\\ \end{aligned}

Determination of UPMP coefficients

\begin{aligned} &r_{1p}=-25\\ \\ \\ \\ &r_{2p}=-3,75\\ \end{aligned}

4. Solution of the canonical equations:

\begin{aligned} &\frac{27EI}{20}\cdot\varphi_{1} -\frac{3EI}{16}\cdot\Delta_{2}=25\\ &-\frac{3EI}{16}\cdot\varphi_{1} +\frac{3EI}{32}\cdot\Delta_{2}=3,75\\ \\ &\varphi_{1} =33,33/EI\\ &\Delta_{2}=106,67/EI\\ \end{aligned}

5. Final diagrams of internal forces in the frame:

Moment diagram [kNm]
\( M_{ost}=M_{p}+M_{1} \cdot \varphi_{1}+M_{2} \cdot \Delta_{2}\\ \)

\begin{aligned} & \mathrm{M}_{\mathrm{A}}=0 \cdot \mathrm{kNm} \\ & \mathrm{M}_{\mathrm{BA}}=\frac{-3 \mathrm{EI}}{4} \cdot \varphi_1+\frac{3 \mathrm{EI}}{16} \cdot \Delta_2=-5 \mathrm{kNm} \\ & \mathrm{M}_{\mathrm{BC}}=\frac{3 \mathrm{EI}}{5} \varphi_1-25=-5 \mathrm{kNm} \\ & \mathrm{M}_{\mathrm{CB}}=0 \cdot \mathrm{kNm} \\ & \mathrm{M}_{\mathrm{CD}}=10 \cdot \mathrm{kNm} \\ & \mathrm{M}_{\mathrm{D}}=\frac{3 \mathrm{EI}}{16} \cdot \Delta_2-5=15 \mathrm{kNm} \end{aligned}

Calculations for the shear force diagram

Element BC:

\begin{aligned} &\sum{M_{C}}=0\\ &{Q_{BC}}\cdot 5-5-8\cdot 5\cdot 2,5=0\\ &{Q_{BC}}=21 \ kN\\ &\sum{Y}=0\\ &-{Q_{CB}}+21-40=0\\ &{Q_{CB}}=-19 \ kN\\ \end{aligned} Coordinate of the extreme moment value in element BC: \begin{aligned} &\frac{x}{21}=\frac{5-x}{19}\\ &x=2,625 \ m\\ \end{aligned} Extreme moment value \( \mathrm{M}_{\mathrm{ex}}=-5-8 \cdot 2.625 \cdot \frac{2.625}{2}+21 \cdot 2.625=22.56 \mathrm{kNm} \)

Element AB:

\begin{aligned} &\sum{M_{A}}=0\\ &{Q_{BA}}\cdot 4+5=0\\ &{Q_{BA}}=-1,25 \ kN\\ &{Q_{AB}}=-1,25\\ \end{aligned}

Element CD:

\begin{aligned} &\sum{M_{D}}=0\\ &{Q_{CD}}\cdot 4+10-15=0\\ &{Q_{CD}}=1,25 \ kN\\ &{Q_{DC}}=1,25 \ kN\\ \end{aligned}

Final diagram of axial forces [kN]

6. Kinematic verification

We assume a determinable system and draw the moment diagram with unit force. (degree of static indeterminacy DSI=1)

\begin{aligned} &\delta_1 =\int \frac{M_{ost}\cdot M_{1}}{EJ} dx= \frac{1}{EI} \left( -\frac{1}{3}\cdot 1\cdot 5\cdot 4 -\frac{1}{3} \cdot 1\cdot 5\cdot 5 +\frac{1}{3} \frac{8\cdot 5^2}{8} \cdot 1 \cdot 5 -\frac{4}{6} \cdot (2\cdot 1\cdot 15 + 10)\right) \approx 0\\ \end{aligned}

7. Static verification

We read the reactions (values and correct signs) from the diagrams of normal forces, shear forces, and bending moments.
Then we write the equations of static equilibrium and check if all the equations are satisfied for the read reactions.

\begin{aligned} &\sum{X}=0 \hspace{2cm} 1,25-1,25=0\\ &\sum{Y}=0 \hspace{2cm} 21+19-4\cdot8=0\\ &\sum{M_{E}}=0\hspace{1.65cm} 21\cdot 2,5-1,25\cdot 4+10-15+1,25\cdot 4-19\cdot 2,5=0\\ \end{aligned}