Cantilever Beam Deflection by Integration: 04-BS-6, May 2016, Q2

Question 2: Cantilever 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 (cantilever), L	6 m
Distributed load, w	20 kN/m, full span
Point load, P	30 kN, at the free end (same direction as w)
Section	W610×125 (CISC): I_x = 985×10⁶ mm⁴
E (steel)	200 GPa
Deflection limit	L/120 = 50 mm

Cantilever beam under UDL plus a tip point load.

Find. The deflection and slope at the free end by double integration, and whether the L/120 serviceability limit is satisfied.

Approach. Measure $x$ from the FIXED end. Write $M(x)$ for the standing beam (both the remaining UDL and the tip load act over the length still to the right of the cut), integrate $EI\,y''=M(x)$ twice, and fix the two constants with the fixed-end conditions $y'(0)=0,\ y(0)=0$.

Moment equation. At a cut $x$ from the wall, the portion of beam to the right (length $L-x$) carries the UDL and the tip load, both producing hogging (negative) moment at the cut: $$M(x) = -\frac{w(L-x)^2}{2} - P(L-x)$$
First integration (slope), $EI\,y'=\int M\,dx + C_1$, with $y'(0)=0\Rightarrow C_1=0$.
Second integration (deflection), $EI\,y=\int EI\,y'\,dx + C_2$, with $y(0)=0\Rightarrow C_2=0$.
Evaluate at the free end $x=L$. Substituting $L=6000$ mm, $w=20$ N/mm, $P=30\,000$ N, $E=200\,000$ MPa, $I=985\times10^6\ \text{mm}^4$: $$\theta_{free} = -6.396\times10^{-3}\ \text{rad}\qquad y_{free} = -27.41\ \text{mm (downward)}$$ (equivalently, by the standard closed-form cantilever results $y=\dfrac{wL^4}{8EI}+\dfrac{PL^3}{3EI}$ and $\theta=\dfrac{wL^3}{6EI}+\dfrac{PL^2}{2EI}$.)
Serviceability check. $$|y_{free}| = 27.41\ \text{mm} < \frac{L}{120}=50.0\ \text{mm}\quad\boxed{\text{beam satisfies the deflection limit}}$$ The deflected shape is a smooth downward-curving cantilever: zero deflection and zero slope at the wall, increasing monotonically to the maximum droop at the free end (no inflection point — the moment stays hogging over the whole span).

Quantity	Value
Slope at free end, θ	6.396×10^-3 rad
Deflection at free end, y	27.41 mm, downward
Allowable (L/120)	50.0 mm
(b) Deflection limit satisfied?	Yes (27.4 mm < 50.0 mm)

Topic: Cantilever beam deflection by the double-integration (moment-curvature) method.

Key relations: $EI\,y''=M(x)$; standard cantilever results $y_{tip}=\dfrac{wL^4}{8EI}+\dfrac{PL^3}{3EI}$, $\theta_{tip}=\dfrac{wL^3}{6EI}+\dfrac{PL^2}{2EI}$.

Why this works: For a cantilever measured from the fixed end, both boundary conditions ($y=0,\ y'=0$) apply at the SAME point ($x=0$), so both constants of integration vanish immediately — no simultaneous equations are needed, unlike a simply-supported beam. Because the UDL and the tip load are both present over the full remaining length at every cut, $M(x)$ is a single smooth quadratic-in-$(L-x)$ expression valid for the whole span.

Common pitfall: Measuring $x$ from the free end instead of the fixed end (this flips which boundary conditions are known at $x=0$ and requires solving simultaneous equations instead), or dropping the sign on $M(x)$ (a cantilever under downward loads hogs — $M$ is negative throughout by the sagging-positive convention).

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 9 (deflection by integration); CISC Handbook of Steel Construction (W610×125 properties).

Question 2 of 8: Cantilever Beam Deflection by Integration

Question 2: Cantilever Beam Deflection by Integration (20 marks)