See theoretical introduction to the Maxwell-Mohr method! It also includes references to more exercises and materials on this topic!
Learn more about integrating with Vereszczagin's method.

Solution of the task:

We number the bars and mark the support reactions to calculate their values. The angle alpha is between bars 2 and 3.

We aim to calculate the normal forces in all truss bars due to the external load.

\begin{array}{lll} \sum{X}=0 & H_A=-10\ kN \\ \sum{M_A}=0 & 10\cdot4+10\cdot7-4V_B=0 &\ V_B=27.5\ kN \\ \sum{M_B}=0 & V_A\cdot4+10\cdot4-20\cdot4+10\cdot3=0 &\ V_A=2.5\ kN \\ \sum{Y}=0 & V_A+V_B-20-10=0 &\ L=P \\ \end{array}

We calculate the forces in all bars using any method, in this example we used the method of nodal equilibrium.

\begin{aligned} &\sum{X}=0 & N_2+10=0 \\ &\sum{Y}=0 &-20- N_1=0 \\ &N_1=-20\ kN \\ &N_2=-10\ kN \\ \end{aligned}

\begin{aligned} &\sum{Y}=0 & N_1+2.5+N_3\cdot\frac{\sqrt{2}}{2}=0 \\ &\sum{X}=0 &-10+N_4 +N_3\cdot\frac{\sqrt{2}}{2}=0 \\ &N_3=\frac{35\sqrt{2}}{2}\ kN \\ &N_4=7.5\ kN \\ \end{aligned}

\begin{array}{lll} &\sum{X}=0 & N_6=N_4=7.5\ kN \\ &\sum{Y}=0 &N_5=-27.5 \ kN\\ \end{array}

\begin{aligned} &\sum{Y}=0 & N_7\cdot0.8-10=0 \\ &\sum{X}=0 &-N_6-N_7\cdot0.6=0 \\ &N_7=12.5\ kN \\ &L=P \\ \end{aligned}

To calculate the displacement, we apply a generalized force corresponding to this displacement. Then we recalculate the support reactions and the forces in all bars due to the virtual load.

\begin{aligned} \sum{M_A}=0 & -4V_B+7=0 &\ V_B=\frac{7}{4}\ kN \\ \sum{M_B}=0 & V_A\cdot4+3=0 &\ V_A=-\frac{3}{4}\ kN \\ \end{aligned}

\begin{aligned} &\sum{Y}=0 & -\frac{3}{4}+N_3\cdot\frac{\sqrt{2}}{2}=0 \\ &\sum{X}=0 &N_4 +N_3\cdot\frac{\sqrt{2}}{2}=0 \\ &N_3=\frac{3\sqrt{2}}{4}\ kN \\ &N_4=-\frac{3}{4}\ kN \\ \end{aligned}

\begin{array}{lll} &\sum{X}=0 &N_6=N_4=-\frac{3}{4}\ kN \\ &\sum{Y}=0 &N_5=-\frac{7}{4} \ kN\\ \end{array}

\begin{array}{ll} \sum{Y}=0 &N_7\cdot0.8-1=0 \\ &N_7=\frac{5}{4} \ kN \end{array}

To simplify the calculations, it is convenient to create a table, especially in a spreadsheet. The sought displacement is the integrated graph of the normal forces due to the external and virtual load. Since these graphs are rectangular, this reduces to multiplying in each row the force in the state P, the unit force, and the length of the bar (integrating two rectangles together), and then summing up the results. The obtained sum divided by the tensile stiffness EA is the sought displacement.

\begin{aligned} &E=205GPa\\ &A=30 cm^2\\ &EA=615\cdot10^6\\ \end{aligned} \begin{aligned} \Delta Y_{C}=\sum\frac{N_P\cdot N_1\cdot l_i}{EA}=\frac{379.7424\cdot10^3}{615\cdot10^6}=0.6174mm\\ \end{aligned}