Question 3 of 8: Vertical deflection at the overhang tip of a symmetric two-span beam (18 marks)

Question 3: Vertical deflection at the overhang tip of a symmetric two-span beam (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. Beam A–B–C–D–E with simple supports at B ($x=2\text{ m}$) and D ($x=10\text{ m}$); overhangs A–B and D–E ($2\text{ m}$ each) carry a 12 kN/m UDL; the central span B–D ($8\text{ m}$) carries a 6 kN/m UDL. C ($x=6\text{ m}$) is the centreline. $EI=2000\text{ kN}\cdot\text{m}^2$ throughout.

Find. The vertical deflection at A.

Approach. By symmetry the two support reactions are equal; the tip deflection is then obtained by the unit-load (virtual-work) method, $\Delta_A=\int \dfrac{mM}{EI}\,dx$, integrating the product of the real moment $M(x)$ and the moment $m(x)$ from a unit downward load at A over all four segments.

1. Reactions by symmetry. Total load $=12(2)+6(8)+12(2)=24+48+24=96\text{ kN}$; by symmetry $R_B=R_D=\boxed{48\text{ kN}}$ each (confirmed by $\sum M_B=0$).
2. Real moment diagram (key values). $M_B=-\tfrac12(12)(2)^2=-24\text{ kN}\cdot\text{m}$ (hogging over the overhang support); $M_C=+24\text{ kN}\cdot\text{m}$ (sagging at midspan, by symmetry $M_D=-24\text{ kN}\cdot\text{m}$ too). The diagram is piecewise quadratic in each segment.
3. Virtual moment diagram. A unit load at A (with the same B, D supports) again gives reactions from statics alone; $m(x)$ is piecewise linear, zero at B and D and growing linearly away from them (largest, negative-going, over the loaded overhangs).
4. Integrate $\int mM\,dx$ segment by segment (Simpson's rule is exact here since $mM$ is at worst cubic on each straight segment): A–B contributes $24.0$, B–C contributes $-32.0$, C–D contributes $-32.0$, D–E contributes $0$ (units kN²·m³). Sum $=24-32-32+0=-40$.
5. Deflection. $\Delta_A=\dfrac{-40}{2000}=-0.020\text{ m}$; the negative sign (relative to the downward unit load) means the tip moves upward. $\boxed{\Delta_A = 20\text{ mm, upward}}$ — confirmed independently by a direct stiffness-method solve of the same beam, which returns the identical 20.0 mm.