Statically indeterminate three bar

This problem is a statically indeterminated truss system (Figure 1).

\(x_1,x_2 : area\ of\ each\ bar,\ {\rm E} : young's\ modulus\)

We cannot determine each reaction forces by static equilibrium due to lack of the number of equations. So remove the centered bar to apply a flexibility method. Then, we can get the displacement, δ (Figure 2).

\(\rm F'_1=P, F'_3=0,\delta=\begin{align}\frac{{\rm P}l}{{\rm E}x_1{\rm cos}\theta}\end{align}\)

Secondly, the redundant force, \(\rm F_2\) loads the bars upward, and find the displacement, \(\delta_\rm F\) (Figure 3).

\(\delta_\rm F=\begin{align}\frac{{\rm F_2}l}{{\rm 2E}x_1{\rm cos}^3\theta}\end{align}\)

Finally, getting back to the original problem, we get to know \(\rm F_1\) and \(\rm F_2\) are tensile and \(\rm F_3\) is compression forces, respectively (Figure 4).

Now, the total downward displacement, \(\delta_2\) by load \(\rm P\) is \(\delta \rm cos\theta-\delta_\rm F\). From the equation, we obtain \(\rm F_2\). The displacement diagram as shown in Figure 5.

\(\delta \rm cos\theta=\delta_2+\delta_F,\ \begin{align}\frac{{\rm P}l}{{\rm E}x_1}=\frac{{\rm F_2}l}{{\rm E}x_2}+\frac{{\rm F_2}l}{{\rm 2E}x_1{\rm cos}^3\theta}\end{align}\)

\(\rm F_2=\begin{align}\frac{2x_2\rm cos^3\theta}{2x_1{\rm cos}^3\theta+ x_2}\end{align}P\)

The other forces \(\rm F_1\) and \(\rm F_3\) can be determined from the force equilibrium in horizontal and vertical directions.

\(\rm F_1cos\theta+F_3cos\theta=Pcos\theta,\ F_1cos\theta+F_2-F_3cos\theta=Pcos\theta\)

\({\rm F_1}=\begin{align}\frac{2x_1\rm cos^3\theta+x_2{\rm(1-cos^2\theta)}}{2x_1{\rm cos}^3\theta+x_2}{\rm P},\ {\rm F_3}=\frac{x_2{\rm cos}^2\theta}{2x_1{\rm cos}^3\theta+x_2}\end{align}{\rm P}\)

The deflections in each direction can be found from the stiffness of bars.

\begin{align}u=\delta {\rm cos}\theta=\frac{{\rm P}l}{{\rm E}x_1},\ v=\delta_2=\frac{{\rm P}l}{{\rm E}}\cdot\frac{2{\rm cos}^3\theta}{2x_1{\rm cos}^3\theta+x_2}\end{align}