Castigliano’s theorem — deflection of an overhanging beam: 16-Civ-B1 Advanced Structural Analysis, December 2018, Q3

Question 3: Castigliano’s theorem — deflection of an overhanging 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. A determinate overhanging beam: a pin at ①, a roller at ② 3 m to the right, and a free tip ③ a further 1 m beyond the roller. A uniformly distributed load runs over the whole 4 m.

Given data
Quantity	Symbol	Value
Span, pin to roller	a	3.0 m
Overhang, roller to tip	b	1.0 m
Uniformly distributed load (full length)	w	6 kN/m
Flexural rigidity, both segments	EI	250 kN·m²

Find. The vertical deflection of the free tip ③, magnitude and direction.

Question 3 — pin, roller and a loaded 1 m overhang. The left support is a plain triangle (a pin), not a built-in end.

Approach. Apply a dummy vertical force Q at the tip, write the bending moment of each segment as a function of Q, differentiate under the integral sign and set Q back to zero: $\delta_{3} = \dfrac{1}{EI}\displaystyle\int M\dfrac{\partial M}{\partial Q}\,\mathrm{d}x$.

Reactions with the dummy load in place. Taking moments about the pin, with the resultant of the distributed load at mid-length, $$R_{2} = \dfrac{w(a+b)^{2}/2 + Q(a+b)}{a}, \qquad R_{1} = w(a+b) + Q - R_{2}$$ At $Q = 0$ these give $R_{1} = 8.00$ kN and $R_{2} = 16.00$ kN, whose sum is the 24 kN total load.
Differentiate the reaction, not just the moment. Because Q sits on the overhang, it changes both reactions: $\partial R_{1}/\partial Q = -b/a = -1/3$. This is the step candidates skip, and skipping it is what turns a correct method into a wrong number.
Moment in the span (x measured from the pin, 0 to a). $$M_{1} = R_{1}x - \dfrac{wx^{2}}{2},\qquad \dfrac{\partial M_{1}}{\partial Q} = -\dfrac{b}{a}x = -\dfrac{x}{3}$$
Moment on the overhang (s measured back from the tip, 0 to b). Only the load beyond the cut matters: $$M_{2} = -Qs - \dfrac{ws^{2}}{2},\qquad \dfrac{\partial M_{2}}{\partial Q} = -s$$
Integrate with Q set to zero. $$\int_{0}^{a} M_{1}\dfrac{\partial M_{1}}{\partial Q}\,\mathrm{d}x = -\dfrac{b}{a}\left(\dfrac{R_{1}a^{3}}{3} - \dfrac{wa^{4}}{8}\right) = -\dfrac{1}{3}\left(72 - 60.75\right) = -3.750$$ $$\int_{0}^{b} M_{2}\dfrac{\partial M_{2}}{\partial Q}\,\mathrm{d}s = \dfrac{wb^{4}}{8} = +0.750$$ The two contributions have opposite signs, and the span term is the larger of the two — that is the whole physics of this question.
Assemble. $$\delta_{3} = \dfrac{-3.750 + 0.750}{250} = -0.01200\ \text{m} \;\Rightarrow\; \boxed{\delta_{3} = 12.0\ \text{mm upward}}$$ The sign is negative against the assumed downward Q, so the tip rises.
Confirm it independently. Superposition gives the same answer in two lines. The span carries its own load plus the hogging moment $wb^{2}/2 = 3.00$ kN·m handed to it by the overhang, so the rotation at the roller is $$\theta_{2} = \dfrac{wa^{3}}{24EI} - \dfrac{(wb^{2}/2)a}{3EI} = 0.02700 - 0.01200 = 0.01500\ \text{rad}$$ The rigid rotation lifts the tip by $\theta_{2}b = 15.0$ mm and the overhang’s own cantilever action drops it by $wb^{4}/8EI = 3.0$ mm, netting 12.0 mm up.

Question 3 — results
Quantity	Value
Reaction at the pin ①	8.00 kN (up)
Reaction at the roller ②	16.00 kN (up)
Moment over the roller	−3.00 kN·m (hog)
Maximum sagging moment in the span	+5.333 kN·m at 1.333 m from the pin
Rotation at the roller	0.01500 rad
Vertical deflection at ③	12.0 mm UPWARD

Question 3 bending moment. The span sags while the overhang hogs; the hog at the roller is what rotates the joint and throws the tip upward.

Topic. Castigliano’s second theorem with a dummy load, on a determinate overhanging beam.

Key relations. $\delta = \partial U/\partial Q = \frac{1}{EI}\int M\,(\partial M/\partial Q)\,\mathrm{d}x$ evaluated at $Q = 0$; simple-span end rotation $wL^{3}/24EI$; end rotation from an applied end moment $ML/3EI$; cantilever tip deflection $wb^{4}/8EI$.

Why this works. Castigliano converts a geometric question into an algebraic one: strain energy is a scalar function of the applied loads, and its derivative with respect to a load is the displacement under that load. The dummy load is a bookkeeping device — it need never be applied, it only has to appear in the moment expressions long enough to be differentiated. The reason the answer is upward is worth internalising: 3 m of distributed load sagging the back span rotates the roller joint far more than 1 m of overhang can pull it down, so the tip is carried up by the rigid rotation of the tangent. A short overhang on a long loaded span almost always rises.

Common pitfall. Treating $R_{1}$ as independent of Q, which sets $\partial M_{1}/\partial Q = 0$ in the span and leaves only the +0.750 overhang term — a 3 mm downward answer instead of 12 mm upward, wrong in both magnitude and direction. The second pitfall is reading the left support as built in: a fixed end makes the beam indeterminate and Castigliano’s theorem in this plain form no longer applies.

Source. Hibbeler, Structural Analysis, 10th ed., §8.5–8.6; Kassimali, Structural Analysis, 6th ed., §7.6.

Question 3 of 9: Castigliano’s theorem — deflection of an overhanging beam

Question 3: Castigliano’s theorem — deflection of an overhanging beam (18 marks)