Question 3 of 8: Vertical Deflection of a Doubly Overhanging Beam

Question 3: Vertical Deflection of a Doubly Overhanging Beam (15 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.

Find. The vertical deflection of point B, with its direction stated.

Approach. The beam is statically determinate, so find the reactions, write the real bending moment $M(x)$, then apply the unit-load (virtual work) method with a unit downward load at B: $\delta_B = \int M m\,\mathrm{d}x / EI$. Because the unit load lies inside the span, its virtual moment $m$ is zero on both overhangs, so only the 6 m span contributes.

1. Find the reactions. Taking moments about A, with distances measured from A and downward loads positive, $$R_C(7-1) = 40(8-1) - 60(1-0) = 280 - 60 = 220$$ $$\boxed{R_C = \tfrac{110}{3} = 36.667\ \text{kN}\ \uparrow,\qquad R_A = 100 - 36.667 = \tfrac{190}{3} = 63.333\ \text{kN}\ \uparrow}$$
2. Write the real bending moment. On the left overhang $M = -60x$, so at A the moment is $M_A = -60.0\ \text{kN}\cdot\text{m}$. Between the supports, $$M(x) = -60x + \tfrac{190}{3}(x-1) = \tfrac{10}{3}x - \tfrac{190}{3}$$ a straight line from $-60.0$ at A through $-50.0$ at B to $-40.0\ \text{kN}\cdot\text{m}$ at C, where the right overhang closes it back to zero. The whole span hogs: there is no sagging anywhere, which is the physical reason the answer will come out upward.
3. Apply the unit load at B and write the virtual moment. With a unit downward load at mid-span of the 6 m span, the virtual reactions are $0.5$ at A and $0.5$ at C, so with $u = x - 1$ measured from A, $$m(u) = \begin{cases} 0.5\,u, & 0 \le u \le 3\\[2pt] 0.5\,(6-u), & 3 \le u \le 6 \end{cases}$$ and $m = 0$ on both overhangs, because a unit load between the supports produces no moment outside them.
4. Integrate the product over the span. With $M(u) = \tfrac{1}{3}(10u - 180)$, $$\begin{aligned} \int_0^{3} M m\,\mathrm{d}u &= \frac{0.5}{3}\int_0^{3}\left(10u^2 - 180u\right)\mathrm{d}u = -120\\ \int_3^{6} M m\,\mathrm{d}u &= \frac{0.5}{3}\int_3^{6}(10u-180)(6-u)\,\mathrm{d}u = -105 \end{aligned}$$ so that $\int M m\,\mathrm{d}x = -225\ \text{kN}^2\cdot\text{m}^3$.
5. Divide by the flexural rigidity. $$\delta_B = \frac{1}{EI}\int M m\,\mathrm{d}x = \frac{-225}{15\,000} = -0.0150\ \text{m}$$ The negative sign means the displacement opposes the assumed downward unit load, so $$\boxed{\delta_B = 15.0\ \text{mm}\ \text{UPWARD}}$$
6. Confirm it independently. Split the hogging moment diagram into a uniform part $M_0 = -50\ \text{kN}\cdot\text{m}$ (the average of $-60$ and $-40$) and a linear antisymmetric part. The antisymmetric part contributes nothing at mid-span, and a simple span carrying equal end moments $M_0$ deflects at mid-span by $M_0L^2/8EI$, giving $$\delta_B = \frac{50(6)^2}{8(15\,000)} = 0.0150\ \text{m} = 15.0\ \text{mm}\ \uparrow$$ which reproduces the integral exactly.