Question 6 of 8: Five-cable cluster by Castigliano

Question 6: Five-cable cluster by Castigliano (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Five cables pinned to a rigid ceiling, arranged symmetrically about the vertical middle cable (cable 1). Cables 2 and 4 make 70° with the ceiling; cables 3 and 5 make 50°. A downward load P acts at the common lower joint.

Find. the force in each of the five cables using Castigliano's first theorem.

Approach. By symmetry the joint moves only vertically by δ. Write the strain energy as a function of δ, and apply Castigliano's first theorem P = ∂U/∂δ.

1. Geometry. Measuring $\alpha$ from the vertical, cable 1 has $\alpha=0$; cables 2,4 have $\alpha=20^\circ$; cables 3,5 have $\alpha=40^\circ$. With the ceiling flat, $L_i=h/\cos\alpha_i$ where $h=L_1=2000$ mm.
2. Elongation and energy. A vertical joint movement $\delta$ stretches cable i by $\delta\cos\alpha_i$, so $U=\sum \tfrac12\frac{EA}{L_i}(\delta\cos\alpha_i)^2=\tfrac12\frac{EA}{h}\delta^2\sum\cos^3\alpha_i.$
3. Castigliano's first theorem. $$P=\frac{\partial U}{\partial\delta}=\frac{EA}{h}\delta\sum\cos^3\alpha_i\ \Rightarrow\ \delta=\frac{P}{(EA/h)\,(1+2\cos^3 20^\circ+2\cos^3 40^\circ)}=0.140\ \text{mm}.$$
4. Cable forces. $N_i=\frac{EA}{h}\delta\cos^2\alpha_i$, giving $$\boxed{N_1=2.81\ \text{kN},\quad N_2=N_4=2.48\ \text{kN},\quad N_3=N_5=1.65\ \text{kN}}.$$

A vertical-equilibrium check, $N_1+2N_2\cos20^\circ+2N_3\cos40^\circ=10.0$ kN, closes exactly, confirming the force set. The steepest cables carry the most load because their stiffness contribution to vertical resistance scales with the cube of the cosine of their inclination.