Beam Deflection by the Method of Integration: 04-BS-6, December 2013, Q1

Question 1: Beam Deflection by the Method of 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.

Simply supported timber beam: L = 4 m, w = 2 kN/m, M₀ = 3 kN·m CCW at A.

Given.

Quantity	Value
Span L	4 m
UDL w	2 kN/m (entire span)
Applied couple M₀ (CCW, at A)	3 kN·m
Cross-section b × h	100 mm × 200 mm
Elastic modulus E	10 GPa

Find. The maximum deflection of the beam, by direct double integration of the moment-curvature relation (no superposition).

Approach. Find the reactions from statics, write M(x) for the full span from a single free body, integrate EI y″ = M(x) twice, apply the two zero-deflection boundary conditions at the supports to fix the constants, then locate the station where the slope vanishes.

Reactions. Summing moments about A (CCW+): $$R_B(4) - w(4)(2) + M_0 = 0 \Rightarrow R_B(4) - 16 + 3 = 0 \Rightarrow R_B = 3.25\ \text{kN}$$ and from $$\Sigma F_y = 0:\ R_A = wL - R_B = 8 - 3.25 = 4.75\ \text{kN}$$
Moment equation. Cutting at x from A and summing moments of everything to the left of the cut (sagging positive): $$M(x) = R_A x - M_0 - \frac{w x^2}{2} = 4.75x - 3 - x^2 \quad [\text{kN}\cdot\text{m},\ x\ \text{in m}]$$ Check: $M(4) = 4.75(4)-3-16 = 0$, matching the roller's zero-moment condition.
Section properties. $$I = \frac{bh^3}{12} = \frac{(100)(200)^3}{12} = 66.667\times10^6\ \text{mm}^4,\qquad EI = (10{,}000)(66.667\times10^6) = 6.667\times10^{11}\ \text{N}\cdot\text{mm}^2$$
First integration (slope). Working in N, mm (x in mm): $$EI\,y'' = M(x) = -x^2 + 4750x - 3\times10^6$$ $$EI\,y' = -\frac{x^3}{3} + 2375x^2 - 3\times10^6 x + C_1$$
Second integration (deflection) and boundary conditions. $$EI\,y = -\frac{x^4}{12} + 791.667x^3 - 1.5\times10^6 x^2 + C_1 x + C_2$$ $y(0)=0 \Rightarrow C_2 = 0$. $y(4000)=0$ gives $$\boxed{C_1 = -1.3333\times10^{9}\ \text{N}\cdot\text{mm}^2}$$
Locate and evaluate the maximum deflection. Setting $EI\,y'=0$ inside the span gives $x = 2195.3\ \text{mm} = 2.195\ \text{m}$ (the only root in $0

Final Results.

Quantity	Value
R_A	4.75 kN ↑
R_B	3.25 kN ↑
I	66.67 × 10⁶ mm⁴
Location of maximum deflection	x = 2.195 m from A
Maximum deflection	5.57 mm, downward

Topic: Beam deflection by direct (double) integration of the moment-curvature equation.

Key relations: $EI\,y'' = M(x)$; two integrations produce $EI y'$ (slope) and $EI y$ (deflection), each with one constant of integration fixed by a geometric boundary condition (here $y=0$ at both simple supports).

Why this works: Euler-Bernoulli beam theory relates curvature directly to bending moment for a linear-elastic, small-deflection beam. Because M(x) is a single smooth polynomial over the whole 4 m span (no discontinuities inside the span — the concentrated couple and both reactions all act at the supports), one moment expression and one pair of constants covers the entire beam; there is no need to match slope/deflection across an internal boundary.

Common pitfall: Forgetting that a maximum deflection occurs where the SLOPE is zero, not at midspan — the applied end couple shifts the point of zero slope away from $x=L/2$. Also, superposition tables (e.g. AISC/Hibbeler appendix cases) are explicitly barred by the question; the constants must come from direct integration of this beam's own M(x).

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 12 (deflection of beams by integration).

Question 1 of 8: Beam Deflection by the Method of Integration

Question 1: Beam Deflection by the Method of Integration (20 marks)