Question 8 of 8: Vertical deflection at C (beam propped by a strut)

Question 8: Vertical deflection at C (beam propped by a strut) (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. Horizontal beam $A(0,0)$–$B(4,0)$ ($2EI$) –$C(6,0)$ ($EI$); pin at $A$; pin at $D(0,3)$; a strut $D$–$B$ (axial, $AE{=}25000$); a UDL over $B$–$C$. Find. $\delta_C$.

Approach. The strut is a two-force member (axial only). Solve the real system (beam + strut) for member moments $M$ and strut force $N$; apply a unit vertical load at $C$ for the virtual $m,n$; then $\delta_C=\int\frac{Mm}{EI}\,dx+\frac{N\,n\,L_{strut}}{AE}$.

1. Real system. Under the UDL $w$ on $B$–$C$, equilibrium of the propped beam gives the strut force $N=-4.17\,w$ (compression) and the corresponding bending moments along $A$–$B$–$C$.
2. Virtual system. A unit downward load at $C$ produces its own bending moments $m$ and strut force $n$.
3. Combine (beam bending + strut axial). Evaluating $\displaystyle \delta_C=\int\frac{Mm}{EI}dx+\frac{NnL}{AE}$ with $EI=3500$ and $AE=25000$ gives the flexibility $\dfrac{\delta_C}{w}=3.42\times10^{-3}\ \text{m per (kN/m)}$, i.e. $\boxed{\delta_C\approx 3.42\,w\ \text{mm (downward)}}$, $w$ in $\text{kN/m}$.

For example, a UDL of $w=10\text{ kN/m}$ would give $\delta_C\approx34.2\text{ mm}$ downward; $w=20\text{ kN/m}$ gives $\approx68.3\text{ mm}$.