The system is doubly kinematically indeterminate, assuming a geometrically determinate system and a displacement plan as shown in the drawing.

From the geometry of the system: \(sin\alpha=\frac{3}{5},\ cos\alpha=\frac{4}{5},\ tg\alpha=\frac{3}{4}\)

The displacements of each rod according to the drawing:

I write the equation for node C:

\(\Phi_C^1+\Phi_C^2=0\)

And the principle of virtual work:

\(\begin{aligned}\\ -W_C^1\cdot u-W_C^2\cdot\frac{3}{5}u-W_B^2\cdot\frac{3}{5}u+P\cdot u=0\rightarrow W_C^1+\frac{3}{5}W_C^2+\frac{3}{5}W_B^2=P\\ \end{aligned}\)

Since rods 1 and 2 are on a Winkler foundation, I write \(\lambda\) for them:

\(\begin{aligned}\\ &\lambda=\sqrt[4]{\frac{kl^4}{EI}}=\sqrt[4]{\frac{0.0064\cdot\frac{EI}{l^4}\cdot l^4}{EI}}=0.2\\ &\lambda_1=0.2\cdot3=0.6\\ &\lambda_2=0.2\cdot5=1\\ \end{aligned}\)

I calculate the internal moments occurring in the equilibrium equations:

\(\begin{aligned}\\ \Phi_C^1&=\frac{EI}{3l}\left(\alpha\left(\lambda_1\right)\cdot\varphi_C-\theta\left(\lambda_1\right)\cdot\frac{u}{3l}\right)=\\ &=\frac{EI}{3l}\left(4.005\cdot\varphi_C-6.027\cdot\frac{u}{3l}\right)=\frac{EI}{l}\left(1.335\cdot\varphi_C-0.67\cdot\frac{u}{l}\right)\\ \Phi_C^2&=\frac{EI}{5l}\left(\alpha^\prime\left(\lambda_2\right)\cdot\varphi_C+\theta^\prime\left(\lambda_2\right)\cdot\frac{\frac{3}{5}u}{5l}-\delta_1^\prime\left(\lambda_2\right)\cdot\frac{\frac{3}{5}u}{5l}\right)=\\ &=\frac{EI}{5l}\left(3.075\cdot\varphi_C+3.338\cdot\frac{3u}{25l}-2.846\cdot\frac{3u}{25l}\right)=\\ &=\frac{EI}{l}\left(0.615\cdot\varphi_C+0.08\cdot\frac{u}{l}-0.068\frac{u}{l}\right)=\frac{EI}{l}\left(0.615\cdot\varphi_C+0.012\frac{u}{l}\right)\\ \end{aligned}\)

Substituting:

\(\begin{aligned}\\ &\frac{EI}{l}\left(1.335\cdot\varphi_C-0.67\cdot\frac{u}{l}\right)+\frac{EI}{l}\left(0.615\cdot\varphi_C+0.012\frac{u}{l}\right)=0\\ &1.95\cdot\varphi_C-0.658\cdot\frac{u}{l}=0\\ \end{aligned}\)

I calculate the internal forces occurring in the equilibrium equations:

\(\begin{aligned}\\ W_C^1&=-\frac{EI}{\left(3l\right)^2}\left(\theta\left(\lambda_1\right)\cdot\varphi_C-\gamma\left(\lambda_1\right)\cdot\frac{u}{3l}\right)=-\frac{EI}{9l^2}\left(6.027\cdot\varphi_C-12.192\cdot\frac{u}{3l}\right)=-\frac{EI}{l^2}\left(0.67\cdot\varphi_C-0.452\cdot\frac{u}{l}\right)\\ W_C^2&=\frac{EI}{\left(5l\right)^2}\left(\theta^\prime\left(\lambda_2\right)\cdot\varphi_C+\gamma^\prime\left(\lambda_2\right)\cdot\frac{\frac{3}{5}u}{5l}-\epsilon^\prime\left(\lambda_2\right)\cdot\frac{\frac{3}{5}u}{5l}\right)=\frac{EI}{25l^2}\left(3.338\cdot\varphi_C+4.925\cdot\frac{3u}{25l}-2.454\cdot\frac{3u}{25l}\right)=\frac{EI}{l^2}\left(0.134\cdot\varphi_C+0.012\cdot\frac{u}{l}\right)\\ W_B^2&=-\frac{EI}{\left(5l\right)^2}\left(\theta^\prime\left(\lambda_2\right)\cdot\varphi_C+\epsilon^\prime\left(\lambda_2\right)\cdot\frac{\frac{3}{5}u}{5l}-\chi^\prime\left(\lambda_2\right)\cdot\frac{\frac{3}{5}u}{5l}\right)=\\ &=-\frac{EI}{25l^2}\left(2.846\cdot\varphi_C+2.454\cdot\frac{3u}{25l}-3.934\cdot\frac{3u}{25l}\right)=-\frac{EI}{l^2}\left(0.114\cdot\varphi_C-0.007\cdot\frac{u}{l}\right)\\ \end{aligned}\)

Substituting:

\(\begin{aligned}\\ &-\frac{EI}{l^2}\left(0.67\cdot\varphi_C-0.452\cdot\frac{u}{l}\right)+\frac{3}{5}\cdot\frac{EI}{l^2}\left(0.134\cdot\varphi_C+0.012\cdot\frac{u}{l}\right)-\frac{3}{5}\cdot\frac{EI}{l^2}\left(0.114\cdot\varphi_C-0.007\cdot\frac{u}{l}\right)=P\\ &-0.658\cdot\varphi_C+0.463\cdot\frac{u}{l}=\frac{Pl^2}{EI}\\ \end{aligned}\)

We have the system of equations:

\(\begin{aligned}\\ &-0.658\cdot\varphi_C+0.463\cdot\frac{u}{l}=\frac{Pl^2}{EI}\\ &1.95\cdot\varphi_C-0.658\cdot\frac{u}{l}=0\\ \end{aligned}\)

After solving, we obtain:

\(\begin{aligned}\\ &\varphi_C=1.4\frac{Pl^2}{EI}\\ &\frac{u}{l}=4.15\frac{Pl^2}{EI}\\ \end{aligned}\)

Substituting into the internal forces:

\(\begin{aligned}\\ W_C^1&=-\frac{EI}{l^2}\left(0.67\cdot\varphi_C-0.452\cdot\frac{u}{l}\right)=-\frac{EI}{l^2}\left(0.67\cdot1.4\frac{Pl^2}{EI}-0.452\cdot4.15\frac{Pl^2}{EI}\right)=0.9378P\\ W_C^2&=\frac{EI}{l^2}\left(0.134\cdot\varphi_C+0.012\cdot\frac{u}{l}\right)=\frac{EI}{l^2}\left(0.134\cdot1.4\frac{Pl^2}{EI}+0.012\cdot4.15\frac{Pl^2}{EI}\right)=0.2374P\\ W_B^2&=-\frac{EI}{l^2}\left(0.114\cdot\varphi_C-0.007\cdot\frac{u}{l}\right)=-\frac{EI}{l^2}\left(0.114\cdot1.4\frac{Pl^2}{EI}-0.007\cdot4.15\frac{Pl^2}{EI}\right)=-0.13055P\\ \end{aligned}\)

From the equilibrium of individual nodes we have:

\(\begin{aligned}\\ &W_B^2+V_B\cdot cos\alpha=0\\ &V_B=-\frac{W_B^2}{cos\alpha}=-\frac{W_B^2}{\frac{4}{5}}=-\frac{5}{4}W_B^2\\ \end{aligned}\)

Substituting:

\(\begin{aligned}\\ &V_B=-\frac{5}{4}\cdot\left(-0.13055P\right)=0.163P\\ \end{aligned}\)

Similarly for nodes A and C:

\(\begin{aligned}\\ &V_A+N_1=0\rightarrow V_A=-N_1\\ \end{aligned}\) \(\begin{aligned}\\ &-N_1cos\alpha+W_C^1sin\alpha-Psin\alpha+W_C^2=0\\ &N_1=WC2cosα+WC1-Psinαcosα=WC2cosα+WC1-Ptgα\\ &V_A=-\frac{W_C^2}{cos\alpha}+\left(P-W_C^1\right)tg\alpha\\ \end{aligned}\)

Substituting:

\(\begin{aligned}\\ &V_A=-\frac{0.2374P}{\frac{4}{5}}+\left(P-0.9378P\right)\cdot\frac{3}{4}=-0.25P\\ \end{aligned}\)

Source:

Mechanika Konstrukcji (MK IPB) exam, June 24, 2017, part-time studies (Warsaw University of Technology, Faculty of Civil Engineering).