Question 3 of 8: Deflection of a non-prismatic beam by virtual work

Question 3: Deflection of a non-prismatic beam by virtual work (18 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. $A$ pin at $x=0$, $C$ roller at $x=6$; $84$ kN at $B$ ($x=3$), $36$ kN at $D$ ($x=9$, cantilever tip); $EI_{AB}=EI$, $EI_{BC}=EI_{CD}=2EI$, $EI=7.5\times10^4$ kN·m². Find. the vertical deflection $\delta_B$.

Approach. Determinate beam → unit-load (virtual-work) method $\displaystyle \delta_B=\int \frac{M\,m}{EI}\,dx$, with the real moment $M$ from the 84/36 kN loads and the virtual moment $m$ from a unit load at $B$.

1. Real reactions and moments. $\sum M_A: 6R_C = 84(3)+36(9) \Rightarrow R_C = 96$ kN, $R_A = 24$ kN. Then $M=24x$ (A–B), $M=24x-84(x-3)$ (B–C), and $M=-36(9-x)$ on the cantilever (0 at B-line references). Key values: $M_B=72$, $M_C=-108$ kN·m.
2. Virtual system. Unit downward load at $B$: $R_A=R_C=0.5$, giving $m=0.5x$ (A–B), $m=3-0.5x$ (B–C), $m=0$ on the cantilever.
3. Integrate with the correct $EI$ per segment. $$\delta_B=\underbrace{\frac{1}{EI}\!\int_0^3 (24x)(0.5x)\,dx}_{=108/EI}+\underbrace{\frac{1}{2EI}\!\int_3^6 M\,m\,dx}_{=13.5/EI}+0 = \frac{121.5}{EI}.$$
4. Evaluate. $\displaystyle \delta_B=\frac{121.5}{7.5\times10^4}=1.62\times10^{-3}\text{ m} = \boxed{1.62\text{ mm}\ (\downarrow)}$.

The deflection is small — about $1/1850$ of the 3 m A–B span — because the stiffer $2EI$ segment covers the region where the virtual moment is largest, and because the cantilever load contributes only indirectly. As a sanity check, ignoring the stiffness step (treating the whole beam as $EI$) would over-predict the deflection by the $13.5/EI$ that the $2EI$ segment saves, i.e. by roughly 11 %, confirming that the non-prismatic modelling matters here.