Vertical Deflection of a Stepped Continuous Beam: 07-Str-A1, December 2017, Q3

Question 3: Vertical Deflection of a Stepped Continuous Beam (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.

Item	Value
Span ①–②	6 m, flexural rigidity $EI_0$
Span ②–③	6 m, flexural rigidity $3EI_0$
Overhang ③–④	3 m, flexural rigidity $3EI_0$
Supports	pin at ① ($x = 0$), roller at ③ ($x = 12$ m), free end at ④ ($x = 15$ m)
Loads	18 kN down at ② ($x = 6$ m); 9 kN down at ④ ($x = 15$ m)
Reference rigidity	$EI_0 = 9000$ kN.m$^2$

Find. The vertical deflection of point ②, with its direction.

Check: the paper prints the reference rigidity as 9000 kN.mm$^2$. Taken literally that is a rigidity some six orders of magnitude too small for a 15 m beam, and the deflection would be nonsense. Read as $EI_0 = 9000$ kN.m$^2$ the answer lands on exactly 33 mm, so the unit is taken as kN.m$^2$ and the answer is quoted in millimetres, which is evidently what the printed “mm” was reaching for.

Approach. The beam is determinate, so use the unit-load method: build the real moment diagram $M$, then the moment diagram $m$ for a unit downward load placed at ②, and evaluate $\delta = \int M m / EI\,dx$ segment by segment, using the correct rigidity in each.

Question 3 — the stepped beam with its real and virtual bending moment diagrams.

Reactions under the real loading. Moments about ① give $$R_3 = \frac{18(6) + 9(15)}{12} = \frac{243}{12} = 20.25\ \text{kN}$$ and vertical equilibrium gives $R_1 = 27 - 20.25 = 6.75$ kN, both upward.
Real moment diagram. Working from the left, $M = 6.75x$ over $0 \le x \le 6$, so $M = +40.5$ kN.m under the 18 kN load. Beyond it $M = 6.75x - 18(x-6) = 108 - 11.25x$, which falls to $-27$ kN.m at the roller; from the right the overhang gives the same value, $-9(3) = -27$ kN.m, which is the check. On the overhang itself $M = -9(15-x)$, closing at zero at the free end.
Virtual moment diagram. Replace the real loads by a single unit downward force at ②. Its reactions are $0.5$ at each support, so $m = 0.5x$ up to ②, giving $m = 3$ m at the load, then $m = 6 - 0.5x$, which reaches zero at the roller. Over the overhang $m = 0$, because a unit load applied between the supports produces no moment beyond the roller. That last observation removes a third of the beam from the integration before any arithmetic is done.
Integrate over span ①–②, rigidity $EI_0$. Here $M = 6.75x$ and $m = 0.5x$, so $$\int_0^6 \frac{Mm}{EI_0}\,dx = \frac{1}{EI_0}\int_0^6 3.375x^2\,dx = \frac{3.375(216)}{3EI_0} = \frac{243}{EI_0}$$
Integrate over span ②–③, rigidity $3EI_0$. Measuring $u = x - 6$ from ②, the diagrams are $M = 40.5 - 11.25u$ and $m = 3 - 0.5u$, so the product integrates to $$\int_0^6 (40.5 - 11.25u)(3 - 0.5u)\,du = 729 - 972 + 405 = 162$$ and dividing by the rigidity $3EI_0$ gives a contribution of $54/EI_0$. The stepped rigidity matters: had this span also been $EI_0$ it would have contributed three times as much.
Add the contributions. The overhang contributes nothing because $m = 0$ there, so $$\delta_2 = \frac{243 + 54}{EI_0} = \frac{297}{9000} = 0.0330\ \text{m} = \boxed{33\ \text{mm downward}}$$ The result is positive, which means it acts in the direction of the assumed unit load — downward, as the sagging first span makes obvious.

Quantity	Value
Reaction at ①	6.75 kN up
Reaction at ③	20.25 kN up
Moment at ② (real)	$+40.5$ kN.m
Moment at ③ (real)	$-27$ kN.m
$\int Mm/EI$, span ①–②	$243/EI_0$
$\int Mm/EI$, span ②–③	$54/EI_0$
Deflection at ②	33 mm downward

Topic. Deflection of a determinate beam of varying flexural rigidity by the unit-load (virtual work) method.

Key relations. $1 \cdot \delta = \int_0^L \dfrac{M m}{EI}\,dx$, where $M$ is the bending moment under the real loads and $m$ the moment under a unit load applied at the point and in the direction of the required deflection. A positive result means the deflection is in the direction of the unit load. For piecewise-constant $EI$ the integral is evaluated segment by segment with the local rigidity in each denominator.

Why this works. The virtual-work statement equates the external work done by the unit load riding through the real displacement with the internal work done by the virtual moments $m$ riding through the real curvatures $M/EI$. Nothing about it requires the two systems to be related, which is precisely its power: the virtual system can be chosen as simply as we like. Here the choice of a unit load at ② makes $m$ vanish over the entire overhang, so a beam that looks like three integrals is really two. The stepped rigidity enters only as a divisor, so a segment made three times stiffer contributes a third as much to the deflection — which is the engineering point the question is testing.

Common pitfall. Using $EI_0$ throughout because the symbol looks like a single constant; the second span and the overhang are at $3EI_0$ and forgetting that inflates the answer to 44 mm. The other frequent slip is integrating the product of two diagrams that change slope at different points in one sweep — here both $M$ and $m$ break at ②, so the two spans must be integrated separately in any case.

Source. Hibbeler, Structural Analysis, 10th ed., §8-1 and §9-4 (method of virtual work applied to beams); Kassimali, 6th ed., §7.4; Leet, Uang and Gilbert, 5th ed., Ch. 9.

Question 3 of 8: Vertical Deflection of a Stepped Continuous Beam

Question 3: Vertical Deflection of a Stepped Continuous Beam (18 marks)