We calculate the degree of static indeterminacy (DSI) and select the basic configuration of the force method - UPMS. DSI = 6-3-1 = 2 In terms of static indeterminacy, this example is a bit specific because we have two vertical reactions along bar AC. This is an axial indeterminacy, which cannot be solved using the force method, but will be solved later when calculating normal forces using the geometric condition. Nevertheless, we will solve this system as a twice indeterminate one, but as we will soon see, one state will turn out to be zero, so it's as if we were calculating a system with DSI=1.

UPMS

We draw unit diagrams and the external load diagram.

State X1=1

State X2=1

State P

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We calculate the coefficients and the free terms of the canonical equations of the force method.

More information about integration can be found in our theoretical introduction to this topic (LINK)
\begin{aligned} & \delta_{11}=\frac{1}{\mathrm{EI}} \cdot(0) \\ & \delta_{22}=\frac{1}{\mathrm{EI}} \cdot\left(\frac{1}{3} \cdot \frac{12}{7} \cdot \frac{12}{7} \cdot 4+\frac{12}{7} \cdot \frac{12}{7} \cdot 8+\frac{1}{3} \cdot \frac{12}{7} \cdot \frac{12}{7} \cdot 3+\frac{1}{3} \cdot 3 \cdot 3 \cdot 3\right)=\frac{1929}{49} \frac{1}{\mathrm{EI}} \\ & \delta_{12}=\frac{1}{\mathrm{EI}} \cdot(0) \quad \delta_{21}=\delta_{12} \\ & \Delta_{1P}=\frac{1}{\mathrm{EI}} \cdot(0) \\ \end{aligned} \( \Delta_{2P}= \)
\begin{aligned} & \Delta_{2 P}=\frac{1}{\mathrm{EI}} \cdot\left(\frac{1}{3} \cdot \frac{12}{7} \cdot 4 \cdot 36+\frac{1}{6} \cdot \frac{12}{7} \cdot 4 \cdot 20-\frac{1}{3} \cdot \frac{12}{7} \cdot 4 \cdot 16+\frac{1}{2} \cdot \frac{12}{7} \cdot 8 \cdot 36-\frac{1}{2} \cdot \frac{12}{7} \cdot 8 \cdot 60-\frac{1}{3} \cdot \frac{12}{7} \cdot 3 \cdot 60\right) \\ & \Delta_{2 P}=-\frac{1392}{7} \frac{1}{\mathrm{EI}} \end{aligned}

We solve the canonical equations and calculate X1 and X2.

\begin{aligned} &\delta_{11}\cdot{x_1}+\delta_{12}\cdot{x_2}+\delta_{1P}=0 \\ &\delta_{21}\cdot{x_1}+\delta_{22}\cdot{x_2}+\delta_{2P}=0 \\ \end{aligned} \begin{aligned} & \left\{\begin{array}{l} 0 \cdot x_1+0 \cdot x_2+0=0 \\ 0 \cdot x_1+\frac{1929}{49 EI} \cdot x_2-\frac{1392}{7 EI}=0 \end{array}\right. \\ & \left\{\begin{array}{l} 0=0 \\ x_2=5.05 \mathrm{kN} \end{array}\right. \end{aligned}

We calculate the final moment using the superposition formula.

\begin{aligned} &M_{FINAL}=M_P+M_1\cdot{X_1}+M_2\cdot{X_2}\\ \end{aligned} We sum up the diagrams

Load Diagram

Calculations for the shear force diagram

\begin{aligned} & \Sigma M_A=0 \\ & -20+44.657+8 \cdot 4 \cdot 2+ \\ & +Q_B \cdot 4=0 \\ & Q_B=-22.164 \mathrm{kN} \\ & \Sigma x=0 \\ & Q_B+8 \cdot 4-Q_A=0 \\ & Q_A=9.836 \mathrm{kN} \end{aligned} Extreme on the moment diagram \begin{aligned} \frac{9.836}{x} & =\frac{22.164}{4-x} \\ x & =1.23 \mathrm{~m} \end{aligned} \begin{aligned} \text{ Max }=-20+Q_A \cdot x-8 \cdot \frac{x^2}{2}=-13.953 \mathrm{kNm} \end{aligned}

\begin{aligned} & \Sigma M_B=0 \\ & Q_C \cdot 3+15.15=0 \\ & Q_C=-5.05 \mathrm{kN} \\ & \Sigma X=0 \\ & Q_B=Q_C \end{aligned}

\begin{aligned} & \Sigma M_B=0 \\ & -44.657-51.343+ \\ & +Q_D \cdot 8=0 \\ & Q_D=12 \mathrm{kN} \\ & \Sigma y=0 \\ & Q_B=Q_D \end{aligned}

\begin{aligned} & \Sigma M_D=0 \\ & 51.343+Q_E \cdot 3=0 \\ & Q_E=-17.114 \mathrm{kN} \\ & \Sigma x=0 \\ & Q_D=Q_E \end{aligned}

Shear Force Diagram

Calculations for the normal force diagram

Node B \begin{aligned} & \Sigma x=0 \\ & -5.05+22.164+ \\ & +N_{B D}=0 \\ & N_{B D}=-17.114 \mathrm{kN} \\ & \Sigma y=0 \\ & N_{B C}-12-N_{A B}=0 \\ & * N_{B C}=12+N_{A B} \\ \end{aligned} Here we have the axial indeterminacy that we mentioned at the beginning

Node D \begin{gathered} \sum x=0 (\text{eq.}) \\ -N_{B D}-17.114=0 \\ 0=0 \\ \Sigma y=0 \\ 12+N_{D E}-12=0 \\ N_{D E}=0 \mathrm{kN} \end{gathered}

Solution of the axial indeterminacy (geometric condition)

The 12kN shear force and its direction tell us in which direction node B will move and which beam will be in tension and which in compression. The elongation of beam BC is equal to the shortening of beam AB - this is the geometric condition. Let's remind ourselves of the formula for the elongation/shortening of a beam: \( \Delta_L = \frac{N\cdot L}{EA} \)

\begin{aligned} & \Delta L_{A B}=\Delta L_{B C} \\ & \frac{-N_{A B} \cdot 4}{E A}=\frac{N_{B C} \cdot 3}{E A} \\ & -4 N_{A B}=3 N_{B C} \\ & N_{A B}=-0.75 N_{B C} \\ & * N_{B C}=12-0.75 N_{B C} \\ & \left\{\begin{array}{l} N_{B C}=6.857 \mathrm{kN} \\ N_{A B}=-5.143 \mathrm{kN} \end{array}\right. \\ & \end{aligned}

Final Normal Force Diagram

Kinematic Verification

\( \delta_i=\int \frac{Max \cdot \overline{M_i}}{EI} d S=0 \)

\begin{aligned} & \delta_2=\frac{1}{\mathrm{EI}} \cdot\left(\begin{array}{l} \frac{1}{3} \cdot 3 \cdot 3 \cdot 15 \cdot 15+\frac{1}{3} \cdot \frac{12}{7} \cdot 4 \cdot 44.657+\frac{1}{6} \cdot \frac{12}{7} \cdot 4 \cdot 20-\frac{1}{3} \cdot \frac{12}{7} \cdot 4 \cdot 16+\frac{1}{2} \cdot \frac{12}{7} \cdot 8 \cdot 44.657- \\ -\frac{1}{2} \cdot \frac{12}{7} \cdot 8 \cdot 51.343-\frac{1}{3} \cdot \frac{12}{7} \cdot 3 \cdot 51.343 \end{array}\right) \\ & \delta_2=-0.055 \frac{1}{\mathrm{EI}} \end{aligned} Relative error
\( \left|\frac{\delta_2}{\Delta_{2 p}}\right|=0.027 \% \quad<1.5 \% \)

Static Verification

We read the reactions (values and correct directions) from the normal, shear force, and bending moment diagrams. Then we write the static equilibrium equations and check if all the equations are satisfied for the read reactions.
\begin{aligned} & \Sigma \mathrm{x}=0 \\ & -5.05-17.114+8 \cdot 4-9.836=-0 \\ & \Sigma \mathrm{y}=0 \\ & 5.143+6.857-12=0 \\ & \Sigma \mathrm{M}_{\mathrm{D}}=0 \\ & -17.114 \cdot 3-5.05 \cdot 3+6.857 \cdot 8+15.15-8 \cdot 4 \cdot 2+9.836 \cdot 4+5.143 \cdot 8-20=0.002 \sim 0 \end{aligned}