Question 2 of 8: Deflection by Integration

Question 2: 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. Same beam, loading and I-section as Question 1 (E = 200 GPa). The bending-moment expression $M(x)=120+10x-\tfrac{5}{6}x^{3}$ kN·m derived there carries over unchanged.

Find. The maximum deflection $y_{max}$, the slope $\theta_A$ at the left support, and whether $|y_{max}|$ clears the L/120 = 50 mm serviceability limit.

Approach. Integrate the Euler–Bernoulli equation $EI\,y''=M(x)$ twice; the two constants follow from the simply-supported boundary conditions $y(0)=y(L)=0$ (note this beam is pinned at BOTH ends, so neither slope is prescribed — only the deflections vanish).

1. Section properties. For the 100×300 mm I-section (15 mm flanges, 20 mm web), $I=93.78\times10^{6}\ \text{mm}^4$, so $EI=18.76\times10^{12}\ \text{N}\!\cdot\!\text{mm}^2$.
2. Integrate twice. With $M(x)$ in N·mm and $x$ in mm, $$EI\,y'=\int M(x)\,dx + C_1, \qquad EI\,y=\int\!\!\int M(x)\,dx\,dx + C_1x + C_2$$ Applying $y(0)=0$ gives $C_2=0$; applying $y(6000\text{ mm})=0$ then solves for $C_1$ (done symbolically, since the closed form is a lengthy quintic/quartic pair).
3. Slope at A. Evaluating $y'(0)=C_1/EI$: $$\boxed{\theta_A=-0.01951\ \text{rad}=-1.1181^{\circ}}$$ (negative — the beam rotates downward-clockwise at A, as expected under a sagging deflection with the couple pinning $M(0^+)$ positive).
4. Maximum deflection. Scanning $y(x)$ over the span (equivalently, solving $y'(x)=0$) locates the extremum at $x=2828$ mm: $$\boxed{y_{max}=-27.99\ \text{mm}}\quad(\text{downward})$$
5. Serviceability check. The allowable limit is $L/120=6000/120=50.0$ mm. Since $|y_{max}|=27.99\ \text{mm} < 50.0\ \text{mm}$, the beam satisfies the deflection limit.