Jib crane: matrix displacement method

Setup: B and C are hinges on the wall, B at the top and C 3 m below it. Member AB is horizontal (4 m) and member AC joins A to C, so AC $=\sqrt{4^2+3^2}=5$ m. Take A at (4, 0), B at (0, 0) and C at (0, -3). A carries a 10 kN vertical load (downward). $E=200$ kN/mm² $=2\times10^8$ kN/m², $A_{AB}=10000$ mm² $=0.01$ m², $A_{AC}=20000$ mm² $=0.02$ m².

Degrees of freedom

Joint A: $u$ (horizontal, right +) and $v$ (vertical, up +). B and C are fixed.

Member stiffness and direction cosines (from A towards the support)

For each bar the contribution to the joint stiffness is $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$:

• AB: $500000\begin{bmatrix}1&0\\0&0\end{bmatrix}$
• AC: $800000\begin{bmatrix}0.64&0.48\\0.48&0.36\end{bmatrix}$

$$[K]=\begin{bmatrix}1012000&384000\\384000&288000\end{bmatrix}\ \text{kN/m}$$

Displacement of joint A

$$\begin{bmatrix}1012000&384000\\384000&288000\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}=\begin{Bmatrix}0\\-10\end{Bmatrix} \;\Rightarrow\; u=0.0267\ \text{mm},\quad v=-0.0703\ \text{mm}$$

A moves 0.027 mm to the right and 0.070 mm downward.

Member forces

Elongation $\delta=-(uc+vs)$ and force $F=k\delta$:

• AB: $\delta=-(u\times(-1))=0.0267$ mm, $F_{AB}=13.333$ kN (tension)
• AC: $\delta=-(u(-0.8)+v(-0.6))=-0.0208$ mm, $F_{AC}=-16.667$ kN (compression)

Check by equilibrium at A: the vertical component of AC is $16.667\times0.6=10.000$ kN, which balances the 10 kN load ✓; its horizontal component $16.667\times0.8=13.333$ kN balances $F_{AB}=13.333$ kN ✓.

Answer: displacement of A: $u=0.027$ mm, $v=-0.070$ mm; $F_{AB}=13.33$ kN (tension), $F_{AC}=-16.67$ kN (compression).