Simply Supported Beam — Deflection by Integration: 04-BS-6, December 2015, Q1

Question 1: Simply Supported Beam — Deflection by Integration (20 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.

Quantity	Value
Span, L	8 m (pin at A, roller at B)
Uniformly distributed load, w	20 kN/m over the full span
Applied couple at B, M₀	500 kN·m, clockwise (confirmed from the source figure)
Section	W250×149: I_x = 259×10⁶ mm⁴ (CISC table)
E (steel)	200 GPa

Beam loading: full-span UDL plus a clockwise couple applied at the right support B (also used by Question 8).

Find. The maximum deflection of the beam by double integration of the moment-curvature relation, and its direction.

Approach. Solve the reactions from global equilibrium (including the applied couple), integrate $EI\,y'' = M(x)$ twice with the two zero-deflection boundary conditions, then locate the deflection extremum where the slope vanishes.

Reactions. Taking moments about A (CCW+) with the clockwise couple entered as $-M_0$: $$R_B L - w\frac{L^2}{2} - M_0 = 0 \;\Rightarrow\; R_B = \frac{wL}{2}+\frac{M_0}{L} = \frac{20(8)}{2}+\frac{500}{8} = 142.5\ \text{kN}$$ $$R_A = wL - R_B = 20(8) - 142.5 = 17.5\ \text{kN}$$
Bending-moment equation. Cutting at $0\le x\le L$ from the left (the couple acts exactly at $x=L$, so it never enters this interior expression): $$M(x) = R_A x - \frac{w x^2}{2}$$
First integration (slope). $$EI\,y' = \int M(x)\,dx = \frac{R_A x^2}{2} - \frac{w x^3}{6} + C_1$$
Second integration (deflection). $$EI\,y = \frac{R_A x^3}{6} - \frac{w x^4}{24} + C_1 x + C_2$$ Boundary conditions $y(0)=0$ and $y(L)=0$ give $C_2=0$ and $$C_1 = \frac{wL^3}{24} - \frac{R_A L^2}{6} = -6.827\times10^{10}\ \text{N}\cdot\text{mm}^2$$
Locate and evaluate the extremum. Setting $y'(x)=0$ gives the deflection extremum at $x = 5.244$ m, where $$\boxed{y_{max} = EI^{-1}\Big[\tfrac{R_A x^3}{6}-\tfrac{w x^4}{24}+C_1 x\Big]_{x=5244\text{mm}} = +20.25\ \text{mm (UPWARD)}}$$ The large 500 kN·m end couple dominates the modest UDL, so the beam cambers upward almost everywhere between the supports (part (b) sketch below), pinched back to zero at both A and B by the boundary conditions.

Part (b): qualitative deflected shape — the beam bows upward, tied to zero at both supports.

Quantity	Value
R_A	17.5 kN
R_B	142.5 kN
Maximum deflection	20.25 mm, upward, at x = 5.244 m from A

Topic: Beam deflection by double integration (moment-curvature method).

Key relations: $EI\,y''=M(x)$; two constants of integration per loaded region, fixed by the support (or continuity) boundary conditions.

Why this works: For small deflections the curvature of the elastic curve equals $M/EI$ at every cross-section. Integrating once recovers the slope (up to a constant fixed by geometry), and integrating again recovers the deflection. Because the applied couple acts exactly at the boundary $x=L$, it enters the problem only through the reactions — the moment expression for the span itself stays a single smooth polynomial in $x$, so only one pair of constants is needed.

Common pitfall: Assuming the deflection extremum sits at midspan by symmetry; an end couple destroys that symmetry, so the true extremum (where $y'=0$) must be solved for explicitly rather than assumed. A second pitfall is forgetting that the couple, applied exactly at a support, does not appear inside $M(x)$ for $x

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 9 (deflection by integration).

Question 1 of 8: Simply Supported Beam — Deflection by Integration

Question 1: Simply Supported Beam — Deflection by Integration (20 marks)