Centre-line deflection of a beam of stepped flexural rigidity: 07-Str-A1, May 2014, Q3

Question 3: Centre-line deflection of a beam of stepped flexural rigidity (15 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. A 12 m simply supported beam carrying two equal point loads placed symmetrically about the centre line, with the flexural rigidity doubled over the middle 8 m.

Question 3 — data
Quantity	Value
Span, point ① to point ⑤	12 m (2 + 4 + 4 + 2)
Supports	pin at ①, roller at ⑤
Loads	30 kN downward at ② ($x = 2$ m) and at ④ ($x = 10$ m)
Flexural rigidity	$EI$ on ①–② and ④–⑤; $2EI$ on ②–④
Reference rigidity	$EI = 1.6 \times 10^{4}$ kN·m²

Find. The vertical deflection of point ③, the centre line at $x = 6$ m. Point ③ carries no support — the symbol on the drawing is a centre-line mark.

Question 3 — real bending moment diagram M (kN·m) and virtual moment diagram m (m) for a unit load at the centre line. The flexural rigidity steps from EI to 2EI at points 2 and 4, so the two diagrams are combined piecewise.

Approach. Use the unit-load (virtual work) method, $\delta = \int M m \, dx / EI$, exploiting the symmetry of both diagrams and the fact that the real moment is constant between the two loads.

Real moment diagram. Symmetry gives $R_{1} = R_{5} = 30$ kN. Hence $M = 30x$ for $0 \le x \le 2$ m, and beyond the first load $$M = 30x - 30(x - 2) = 60\ \text{kN}\cdot\text{m}$$ so the moment is constant at 60 kN·m over the whole middle 8 m, which is exactly the region of doubled rigidity.
Virtual moment diagram. Remove the real loads and apply a unit downward force at the centre line. Its reactions are $0.5$ each, so $$m = 0.5x \ (0 \le x \le 6\ \text{m}), \qquad m = 0.5(12 - x)\ (6 \le x \le 12\ \text{m}).$$
Assemble the integral. Both $M$ and $m$ are symmetric about mid-span, so integrate the left half and double: $$\delta = \frac{2}{EI}\left[\int_{0}^{2}\frac{(30x)(0.5x)}{1}\,dx + \int_{2}^{6}\frac{(60)(0.5x)}{2}\,dx\right]$$ Note that the rigidity ratio enters as a divisor inside each piece, which is the whole reason for splitting the integral at $x = 2$ m.
Evaluate the two pieces. $$\int_{0}^{2} 15x^{2}\,dx = 5x^{3}\Big|_{0}^{2} = 40\ \text{kN}\cdot\text{m}^{3}$$ $$\int_{2}^{6} 15x\,dx = 7.5x^{2}\Big|_{2}^{6} = 7.5(36 - 4) = 240\ \text{kN}\cdot\text{m}^{3}$$ so the bracket is $40 + 240 = 280$ kN·m³.
Deflection. Doubling and dividing by the reference rigidity, $$\delta_{3} = \frac{2(280)}{1.6 \times 10^{4}} = \frac{560}{1.6 \times 10^{4}} = \boxed{0.0350\ \text{m} = 35.0\ \text{mm} \downarrow}$$ The positive sign means the deflection is in the direction of the applied unit load, that is downward.
Sanity check. Had the beam been of uniform $EI$ throughout, the same integral would give $2(40 + 480)/EI = 65.0$ mm. Stiffening only the middle 8 m therefore removes almost half of the deflection, which is the expected result because the middle of the span is where the product $Mm$ is largest.

Question 3 — results
Quantity	Value
Support reactions	30.0 kN at each end
Real moment on the middle 8 m	60.0 kN·m (constant)
Virtual moment at the centre line	3.00 m
$\int M m \,(EI_{\text{ref}}/EI)\,dx$	560 kN·m³
Deflection at point ③	35.0 mm downward

Topic. Deflection of a non-prismatic beam by the unit-load (virtual work) method.

Key relations. $1 \cdot \delta = \displaystyle\int_{0}^{L} \frac{M m}{EI}\,dx$; for a stepped beam write $EI(x) = \alpha EI_{\text{ref}}$ and carry $\alpha$ as a divisor in each piece.

Why this works. The unit-load method is a statement of the principle of virtual forces: a statically admissible virtual force system does virtual work on the real deformations, and for a beam that internal work is $\int M m \, dx / EI$. Because the virtual system need only be in equilibrium, any convenient statically determinate release can be used. The method handles a stepped rigidity with no extra machinery at all — the step just adds an integration limit — whereas double integration would need two constants per segment and four continuity conditions.

Common pitfall. Treating point ③ as a support because three circled numbers on the drawing sit near supports. It is the centre-line mark; a support there would make the deflection zero and the question meaningless. The second pitfall is multiplying by the rigidity ratio instead of dividing by it, which stiffens the wrong region and yields roughly 60 mm.

Source. Hibbeler, Structural Analysis 10th ed., §8.4 and §9.1–9.4; Kassimali, Structural Analysis 6th ed., §7.4.

Question 3 of 8: Centre-line deflection of a beam of stepped flexural rigidity

Question 3: Centre-line deflection of a beam of stepped flexural rigidity (15 marks)