Vertical deflection at B of a beam with stepped stiffness: 07-Str-A1, December 2014, Q3

Question 3: Vertical deflection at B of a beam with stepped stiffness (18 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 15 m beam ABCD, pinned at A and on a roller at C, with the stiffness stepping up from $EI$ to $2EI$ at the 84 kN load point.

Given data
Item	Value
Pin support A	$x = 0$
Point load 84 kN downward	$x = 3$ m
Point B (deflection required)	$x = 6$ m
Roller support C	$x = 12$ m
Point load 36 kN downward at the free end D	$x = 15$ m
Flexural rigidity	$EI$ over $0 \le x \le 3$ m, $2EI$ over $3 \le x \le 15$ m
Reference value	$EI = 7.5 \times 10^{4}\ \text{kN}\cdot\text{m}^{2}$

Find. The vertical deflection of point B, in millimetres, stating its direction.

Beam for Question 3 with the real bending moment M and the virtual moment m produced by a unit downward load at B. Note that m vanishes beyond the roller, so the 3 m overhang contributes nothing.

Approach. Use the unit-load (virtual work) method, $1 \cdot \Delta = \int M m / EI\ \mathrm{d}x$, evaluating the integral piece by piece so that the change of rigidity at $x = 3$ m and the kinks in $m$ at B and C are respected.

Reactions in the real system. Moments about A give $$R_{C} = \frac{84(3) + 36(15)}{12} = \frac{252 + 540}{12} = \boxed{66\ \text{kN}}, \qquad R_{A} = 84 + 36 - 66 = \boxed{54\ \text{kN}}$$ Both act upwards; the small reaction at A reflects how much of the tip load is carried by the overhang lever.
Real bending moment. Working from the left, $$M = 54x \quad (0 \le x \le 3), \qquad M = 54x - 84(x - 3) = 252 - 30x \quad (3 \le x \le 12)$$ and $M = -36(15 - x)$ over the overhang. The peak sagging moment is $M = 162\ \text{kN}\cdot\text{m}$ under the 84 kN load, and the hogging moment over the roller is $-108\ \text{kN}\cdot\text{m}$, which equals $-36(3)$ read from the right — a one-line check on the whole diagram.
Virtual system. Remove the real loads and apply a unit downward load at B. Its reactions are $0.5$ at A and $0.5$ at C, so $$m = 0.5x \quad (0 \le x \le 6), \qquad m = 6 - 0.5x \quad (6 \le x \le 12)$$ and $m = 0$ over CD, because no force in the virtual system lies beyond C. The peak virtual ordinate is $m = 3.0$ m at B. Discarding the overhang at this point saves a third of the arithmetic.
Integrate segment by segment. Writing $EI$ for the reference rigidity and carrying the relative factor explicitly, $$\int_{0}^{3} \frac{(54x)(0.5x)}{EI}\,\mathrm{d}x = \frac{243}{EI}, \qquad \int_{3}^{6} \frac{(252-30x)(0.5x)}{2EI}\,\mathrm{d}x = \frac{378}{EI}$$ $$\int_{6}^{12} \frac{(252-30x)(6-0.5x)}{2EI}\,\mathrm{d}x = \frac{54}{EI}, \qquad \int_{12}^{15} = 0$$ Note the third integral is small because $M$ changes sign at $x = 8.4$ m while $m$ stays positive, so the two halves of that segment very nearly cancel.
Add up and substitute. Summing the four contributions, $$\Delta_{B} = \frac{243 + 378 + 54}{EI} = \frac{675}{EI} = \frac{675}{7.5 \times 10^{4}} = \boxed{0.0090\ \text{m} = 9.0\ \text{mm downward}}$$ The result is positive in the sense of the unit load, which was applied downwards, so B deflects downwards. As a sanity check, 9 mm over a 12 m span is about $L/1300$ — entirely reasonable for a beam of this stiffness under these loads.

Question 3 — results
Quantity	Value
Reaction at A	54 kN upward
Reaction at C	66 kN upward
Maximum sagging moment (at the 84 kN load)	$162\ \text{kN}\cdot\text{m}$
Hogging moment over C	$-108\ \text{kN}\cdot\text{m}$
$\int Mm/EI\ \mathrm{d}x$ (relative to $EI$)	$675\ \text{kN}\cdot\text{m}^{3}$
Vertical deflection at B	9.0 mm downward

Topic. Deflection of a determinate beam of non-uniform flexural rigidity by the unit-load method.

Key relations. $1 \cdot \Delta = \displaystyle\int_{0}^{L} \frac{M m}{EI}\,\mathrm{d}x$, where $M$ is the bending moment under the real loads and $m$ the moment under a unit load applied at, and in the direction of, the deflection sought. A positive result means the point moves in the direction of the unit load.

Why this works. The expression is a statement of the principle of virtual forces: the external virtual work done by the unit load moving through the real displacement equals the internal virtual work done by the virtual moments acting through the real curvatures $M/EI$. Because $EI$ appears inside the integral, a stepped beam is handled simply by breaking the integral at every change of section, and there is no need to construct a transformed section or an equivalent uniform beam. Splitting at the kinks in $m$ as well keeps every sub-integral a polynomial that can be evaluated in closed form, which is what makes the method practical under exam conditions.

Common pitfall. Integrating across the step in $EI$, or across a kink in either diagram, with a single product formula or a moment-area table entry — both are only valid over a segment where $M$ and $m$ are smooth and $EI$ is constant. A second frequent slip is forgetting that the virtual moment is zero beyond the support nearest the free end when the unit load lies inside the span, and grinding out an integral over the overhang that must come to nothing.

Source. Hibbeler, Structural Analysis, 10th ed., §8.4 and §9.4–9.6 (virtual work for beams and frames); Kassimali, Structural Analysis, 6th ed., §7.3–7.4.

Question 3 of 8: Vertical deflection at B of a beam with stepped stiffness

Question 3: Vertical deflection at B of a beam with stepped stiffness (18 marks)