Simply-Supported Beam Deflection by Double Integration: 04-BS-6, Undated paper, Q2

Question 2: Simply-Supported Beam Deflection by Double Integration

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 $L$ (pin at A, roller at B)	4 m
UDL $w$	10 kN/m, from $x=1$ m to $x=4$ m (ends at B)
Section	W150×30: $I_x = 17.1\times10^{6}\text{ mm}^4$, $S_x=218\times10^3\text{ mm}^3$
$E$	200 GPa
Allowable $\sigma$, $\tau$	240 MPa, 60 MPa

Find. The maximum deflection of the beam, by direct (Method-of-Integration) solution of $EI\,v''=M(x)$ — superposition is explicitly disallowed.

Simple span 4 m; 10 kN/m UDL over the last 3 m (x=1 to x=4).

Approach. Find reactions, write $M(x)$ piecewise for the two regions ($0\le x\le1$, $1\le x\le4$), integrate $EIv''=M(x)$ twice per region, apply the four conditions ($v(0)=0$, $v(4)=0$, slope/deflection continuity at $x=1$), then locate where $v'=0$.

Reactions. $\sum M_A=0$: $R_B(4)=10(3)(2.5)=75\Rightarrow R_B=18.75$ kN; $R_A=30-18.75=11.25$ kN.
Moment equations. Region 1 ($0\le x\le1$): $M_1=11.25x$. Region 2 ($1\le x\le4$): $M_2=11.25x-5(x-1)^2$ (kN·m).
Integrate twice per region, match at $x=1$, apply BCs. With $EI = (200\times10^9)(17.1\times10^{-6})=3.42\times10^6\text{ N}\!\cdot\!\text{m}^2$, solving the 4 constants gives $$EI\,v_1'(x)=5625x^2-21562.5,\qquad EI\,v_1(x)=1875x^3-21562.5x$$ $$EI\,v_2'(x)=5625x^2-\tfrac{5000}{3}(x-1)^3-19895.83,\qquad EI\,v_2(x)=1875x^3-\tfrac{1250}{3}(x-1)^4-19895.83x-416.67$$ (all in N, m).
Locate and evaluate the maximum. Setting $v_2'(x)=0$ inside the loaded region gives $x=2.0416$ m. Substituting, $$\boxed{v_{max}=8.35\text{ mm (downward), at } x=2.04\text{ m from A}}$$
Design check (bonus). $V=0$ at $x=2.125$ m gives $M_{max}=17.58$ kN·m, so $\sigma_{max}=M_{max}/S_x=17.58\times10^6/218\times10^3=80.6$ MPa $< 240$ MPa allowable — the section is governed by deflection/serviceability here, not by strength.

Result	Value
$R_A$, $R_B$	11.25 kN, 18.75 kN
Maximum deflection	8.35 mm, at $x=2.04$ m
$M_{max}$ (design check)	17.58 kN·m → $\sigma=80.6$ MPa (adequate)

Topic: Beam deflection by the double-integration (Method of Integration) technique.

Key relations: $EI\,v''(x)=M(x)$; continuity of slope $v'$ and deflection $v$ across a load discontinuity; boundary conditions $v=0$ at simple supports.

Why this works: Euler–Bernoulli beam theory relates curvature directly to bending moment. Integrating twice recovers slope then deflection; because the load is piecewise (starts at $x=1$), $M(x)$ is a different polynomial in each region, and the four integration constants are fixed by matching the two regions at their boundary (continuity) and by the two support conditions.

Common pitfall: Using one set of constants for the whole beam (ignoring the load discontinuity), or forgetting that the maximum deflection occurs where $v'=0$ — not necessarily at the same $x$ as $M_{max}$.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 8 §8.2 (Slope and Displacement by Integration).

Question 2 of 8: Simply-Supported Beam Deflection by Double Integration

Question 2: Simply-Supported Beam Deflection by Double Integration