Vertical deflection at point 2 of a two-span beam with an overhang (18 marks): 24-Bld-A1 Elementary Structural Analysis, December 2017, Q3

Question 3: Vertical deflection at point 2 of a two-span beam with an overhang (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.

Check: the source prints "EI₀ = 9000 kN.mm²" — for an 8 m span, kN·mm² is off by a factor of 10⁶ and gives an absurd (kilometre-scale) deflection. This solution adopts the dimensionally sensible

EI_0=9000\text{ kN}\cdot\text{m}^2

, which also reproduces a clean, exam-scale answer.

Given. Beam over 4 points; span 1–2 = 6 m at $$$EI_0$$$ ; span 2–3 = 6 m and span 3–4 = 3 m (overhang) both at $$$3EI_0$$$ ; pin at 1, roller at 3; 18 kN downward at point 2; 9 kN downward at the tip, point 4.

Find. The vertical deflection at point 2.

Points 1–4, EI regions, and the 18 kN / 9 kN loads.

Approach. The beam is statically determinate (pin + roller = 3 reactions, single span with an overhang). Find reactions, then apply the unit-load (virtual work) method with a unit downward load at point 2, integrating $$$M m/EI$$$ piecewise over the three EI regions.

Reactions. $\Sigma M_1=0:\ R_3(12)=18(6)+9(15)\Rightarrow R_3=\dfrac{108+135}{12}=20.25\text{ kN}$ $R_1=18+9-20.25=6.75\text{ kN}$
Real moment, M(x) (x from point 1): $M=6.75x\ (0\le x\le6);\quad M=6.75x-18(x-6)\ (6\le x\le12);\quad M=6.75x-18(x-6)+20.25(x-12)\ (12\le x\le15)$
Virtual system — unit load down at point 2 (x=6): $R_{1v}(12)=1(6)\Rightarrow R_{1v}=0.5,\quad R_{3v}=0.5$ $m=0.5x\ (0\le x\le6);\quad m=0.5x-(x-6)\ (6\le x\le12);\quad m=0.5x-(x-6)+0.5(x-12)\ (12\le x\le15)$
Integrate $\Delta_2=\int \dfrac{Mm}{EI}\,dx$ over the three segments: $\int_0^6 \dfrac{Mm}{EI_0}dx=\dfrac{243}{EI_0},\qquad \int_6^{12}\dfrac{Mm}{3EI_0}dx=\dfrac{54}{EI_0},\qquad \int_{12}^{15}\dfrac{Mm}{3EI_0}dx=0$ (the third integral vanishes because $m\equiv0$ beyond the roller for a unit load at point 2). $\Delta_2=\dfrac{243+54}{EI_0}=\dfrac{297}{EI_0}=\dfrac{297}{9000}=\boxed{0.033\text{ m}=33.0\text{ mm}\ \downarrow}$

Quantity	Value
R_1 (pin)	6.75 kN ↑
R_3 (roller)	20.25 kN ↑
Deflection at point 2	33.0 mm, downward

Concept & Theory

Topic: Unit-load (virtual work) method for beam deflection with piecewise-varying EI.

Key relations: $1\cdot\Delta=\int \dfrac{M(x)\,m(x)}{EI(x)}\,dx$ , summed member-by-member wherever EI changes.

Why this works: the virtual-work identity equates external virtual work (unit load times the real deflection) to internal virtual work (virtual moment times real curvature, integrated along the beam) — it needs no prior knowledge of the deflected shape and handles a stepped EI naturally by splitting the integral at each change.

Common pitfall: using a single EI for the whole span, or forgetting that the virtual moment for a unit load applied strictly between two supports is zero beyond the far support (no virtual reaction develops there once the real load position is inside the span).

Source: Hibbeler, Structural Analysis, Ch. 8–9 (virtual work for deflections).

Question 3 of 8: Vertical deflection at point 2 of a two-span beam with an overhang (18 marks)

Question 3: Vertical deflection at point 2 of a two-span beam with an overhang (18 marks)