Question 7 of 8: Deflection of a tie-rod-propped frame

Question 7: Deflection of a tie-rod-propped frame (22 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. A horizontal beam 1–2–3–4 (joints at $0,3,6,9$ m) rigidly joined at the pin 1 to an $8$ m vertical beam 1–5; a tie rod from 5 $(0,8)$ to 3 $(6,0)$ — a $6$–$8$–$10$ triangle, so $L_{tie}=10$ m. $EI=36{,}000$ kN·m² (beams), $EA=25{,}000$ kN (rod). Load $48$ kN down at point 2. Find. $\delta_{4}$, then $\delta_2$ for the moved load.

Approach. The tie rod is one redundant. Solve the propped structure by the force method (compatibility at the rod), then apply a unit vertical load at point 4 and combine by virtual work, $\delta=\displaystyle\int\frac{Mm}{EI}\,dx+\frac{N n L}{EA}$, including the rod’s axial flexibility.

1. Redundant. Releasing the rod and enforcing zero relative movement along its axis gives a rod tension of about $22.2$ kN under the $48$ kN load; this prop moment shapes the real bending diagram $M$.
2. Unit-load system. Apply a unit downward load at point 4 and find $m$ (beam) and $n$ (rod) for the same released structure.
3. Combine. Evaluating $\displaystyle\int\frac{Mm}{EI}\,dx+\frac{NnL}{EA}$ over both beams and the rod gives $\delta_4=\boxed{10.8\ \text{mm}\ \downarrow}$.
4. (b) Reciprocity. By Maxwell’s reciprocal theorem the deflection at 2 caused by a load at 4 equals the deflection at 4 caused by the same load at 2. Hence $\delta_2=\boxed{10.8\ \text{mm}\ \downarrow}$ — identical, without any further analysis.