Truss analysis by displacement method

Setup: the joint O hangs 5 m below the horizontal line of supports and is held by four bars. Measured from the +x axis (towards the support), the bars make angles $120^\circ$ (bar 1), $90^\circ$ (bar 2), $60^\circ$ (bar 3) and $45^\circ$ (bar 4). This reading of the figure is assumed. Loads at O: $P_x=50$ kN (to the right) and $P_y=-100$ kN (downward). $E=2\times10^5$ MPa $=2\times10^8$ kN/m², $A=8$ cm² $=8\times10^{-4}$ m², so $EA=1.6\times10^5$ kN. The degrees of freedom are the joint displacements $u$ (right) and $v$ (up). All support joints are fixed.

Bar data ($L=5/\sin\alpha$, $k=EA/L$, direction cosines $c=\cos\alpha$, $s=\sin\alpha$ from O towards the support)

Stiffness matrix of the joint

For one bar the contribution to the joint is $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$. Summing the four bars:

$$[K]=\begin{bmatrix}25170.1&11313.7\\11313.7&84882.9\end{bmatrix}\ \text{kN/m}$$

Solution of $[K]\{d\}=\{P\}$

$$\begin{bmatrix}25170.1&11313.7\\11313.7&84882.9\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}50\\-100\end{Bmatrix} \;\Rightarrow\; u=2.6764\ \text{mm},\quad v=-1.5348\ \text{mm}$$

Bar forces

Elongation of a bar: $\delta=-(uc+vs)$ (the joint moving away from the support lengthens it), force $F=k\delta$ (tension +).

Equilibrium check at O: $\sum F_x=0.000$, $\sum F_y=0.000$ ✓.

Answer: joint displacement $u=2.676$ mm, $v=-1.535$ mm; bar forces $F_1=73.92$, $F_2=49.11$, $F_3=-0.25$, $F_4=-18.26$ kN (+ tension).