Maximum Deflection of a Beam Under Triangular Load: 04-BS-6, May 2018, Q3

Question 3: Maximum Deflection of a Beam Under Triangular Load (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. Span $L=9\text{ m}$, triangular load $w(x)=w_0 x/L$ with $w_0=6\text{ kN/m}$ at the roller end (zero at the pin end); $E=200\text{ GPa}$; symmetric I-section: flanges $100\times10\text{ mm}$, web $10\text{ mm}$ thick, overall depth $300\text{ mm}$.

Find. The location $x_{\max}$ (measured from the zero-load end A) and value $y_{\max}$ of the maximum deflection, by double integration of the elastic curve.

Triangular load: 0 at pin A rising to 6 kN/m at roller B, span 9 m.

Symmetric I-beam cross-section: 100×10 mm flanges, 10 mm web, 300 mm overall depth.

Deflected shape (exaggerated): maximum deflection 21.3 mm at x=4.67 m from A.

Approach. Find the reactions and M(x) for the triangular load, then integrate $EI\,y^{\prime\prime}=M(x)$ twice, apply the two boundary conditions $y(0)=y(L)=0$, and locate the deflection extremum where the slope vanishes.

Section property. $I=2\left[\dfrac{b_ft_f^3}{12}+b_ft_f\left(\tfrac{d}{2}-\tfrac{t_f}{2}\right)^2\right]+\dfrac{t_w(d-2t_f)^3}{12}=2(8{,}333+21{,}025{,}000)+18{,}293{,}333=\boxed{60.36\times10^{6}\text{ mm}^4}$.
Reactions and moment. For $w(x)=w_0x/L$, $R_A=\dfrac{w_0L}{6}=9\text{ kN}$, $R_B=\dfrac{w_0L}{3}=18\text{ kN}$ (sum $=27\text{ kN}$, the triangle's area, checks out). Integrating the load, $M(x)=R_Ax-\dfrac{w_0x^3}{6L}$, which correctly gives $M(0)=M(L)=0$.
First integration (slope). $EIy^{\prime}=\dfrac{R_Ax^2}{2}-\dfrac{w_0x^4}{24L}+C_1$.
Second integration (deflection) and boundary conditions. $EIy=\dfrac{R_Ax^3}{6}-\dfrac{w_0x^5}{120L}+C_1x+C_2$. $y(0)=0\Rightarrow C_2=0$; $y(L)=0$ then fixes $C_1$. Solving numerically gives $C_1=-85.05\times10^{9}\text{ N}\cdot\text{mm}^2$.
Locate and evaluate the extremum. Setting $y^{\prime}(x)=0$ and solving the resulting quartic numerically gives $x_{\max}=\boxed{4.674\text{ m from A}}$ (the zero-load end — NOT mid-span, since the load is skewed toward B). Substituting back, $y_{\max}=\boxed{-21.3\text{ mm}}$ (downward), using $EI=200{,}000\times60.36\times10^6=1.207\times10^{13}\text{ N}\cdot\text{mm}^2$.

Final results
Quantity	Value
I (section)	60.36 × 10⁶ mm⁴
x_max (from A)	4.674 m
y_max	-21.3 mm (downward)

Topic: Beam deflection by double integration (Macaulay-free, single-region load)

Key relations: $EIy^{\prime\prime}=M(x)$; integrate twice; boundary conditions $y(0)=y(L)=0$ for a simply supported beam.

Why this works: Because the load is a single continuous function over the whole span, one moment expression covers the beam and double integration directly yields the elastic curve; the two constants of integration are fixed by the two zero-deflection boundary conditions at the supports, not by guessing.

Common pitfall: Assuming the maximum deflection occurs at mid-span; for an asymmetric (triangular) load the extremum shifts toward the heavier-loaded end, and its location must be solved for by setting the slope to zero, not assumed.

Source: Hibbeler, Mechanics of Materials, 10th ed., §9.2 The Elastic Curve (Method of Integration)

Question 3 of 8: Maximum Deflection of a Beam Under Triangular Load

Question 3: Maximum Deflection of a Beam Under Triangular Load (20 marks)