Question 3 of 9: Castigliano's theorem — deflection at an internal hinge

Question 3: Castigliano's theorem — deflection at an internal hinge (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 beam ABCD built in at A and carried on a roller at D, with an internal hinge at C. Each bay is 3 m long. A uniformly distributed load of 8 kN/m acts over AB and a second uniformly distributed load of 8 kN/m acts over CD; the middle bay BC is unloaded.

Find. The vertical deflection of point C, the hinge, in millimetres, together with its direction.

Approach. The hinge makes the structure determinate, so split it there: CD is a simply supported span whose reaction at the hinge is a known force on the remaining cantilever AC, and Castigliano's first theorem is then applied to that cantilever alone with a dummy force $Q$ at C.

1. Check that the structure is determinate. The reactions are three at A and one at D, four in all, against three equations of equilibrium plus the condition $M_C = 0$ supplied by the hinge. Four equations for four unknowns, so no redundant exists and the internal forces follow from statics alone — which is what makes a single Castigliano integral sufficient.
2. Isolate span CD and find the force it hands to the hinge. CD spans 3 m between the hinge at C and the roller at D and carries 8 kN/m, with zero moment at both ends, so it behaves as a simply supported span. Taking moments about C, $$R_D = \frac{w\,a^{2}/2}{a} = \frac{w\,a}{2} = \frac{8 \times 3}{2} = \boxed{12\ \text{kN}}$$ and by vertical equilibrium the hinge delivers the other half, $V_C = 8(3) - 12 = 12$ kN, downwards on the cantilever AC.
3. Reduce the problem to a cantilever. Everything to the right of the hinge is now replaced by that 12 kN tip force, so AC is a 6 m cantilever built in at A, carrying 8 kN/m over its first 3 m (the AB bay) and a 12 kN point load at its tip. Its base reactions follow immediately: $$R_A = 8(3) + 12 = 36\ \text{kN}, \qquad M_A = 8(3)(1.5) + 12(6) = 36 + 72 = \boxed{108\ \text{kN}\cdot\text{m}}$$ (hogging), which will serve later as an arithmetic check.
4. Introduce the dummy load and write the moment function. Add a vertical force $Q$ at C in the direction of the required deflection and measure $s$ from C towards A. Only the tip forces act over the first 3 m; beyond that the UDL enters: $$M(s) = -(12 + Q)\,s \qquad (0 \le s \le 3)$$ $$M(s) = -(12 + Q)\,s - \tfrac{1}{2}w\,(s-3)^{2} \qquad (3 \le s \le 6)$$ Note that span CD stores strain energy but its moments do not contain $Q$: a force applied at the hinge is carried straight into the cantilever and does not load the simply supported span, so that segment contributes nothing to the derivative.
5. Apply Castigliano's first theorem. With $\partial M/\partial Q = -s$ throughout and $Q$ set to zero after differentiating, $$\Delta_C = \frac{\partial U}{\partial Q}\bigg|_{Q=0} = \frac{1}{EI}\int_{0}^{6} M\,\frac{\partial M}{\partial Q}\,ds = \frac{1}{EI}\left[\int_{0}^{6} 12s\cdot s\,ds + \int_{3}^{6} 4(s-3)^{2}\,s\,ds\right]$$ Both integrands are positive, which already tells us the deflection is downwards, in the direction assumed for $Q$.
6. Evaluate the two integrals. The tip-load term is the familiar cantilever result: $$\int_{0}^{6} 12s^{2}\,ds = 12\,\frac{6^{3}}{3} = 864 \qquad\Longleftrightarrow\qquad \frac{P L^{3}}{3} = \frac{12(216)}{3} = 864$$ For the distributed term substitute $u = s - 3$: $$\int_{3}^{6} 4(s-3)^{2}s\,ds = 4\int_{0}^{3}u^{2}(u+3)\,du = 4\left[\frac{u^{4}}{4} + u^{3}\right]_{0}^{3} = 4(20.25 + 27) = 189$$ The same figure follows from the standard partial-UDL cantilever formula $w a^{3}(4L - a)/24 = 8(27)(24 - 3)/24 = 189$.
7. Add the two contributions and put in the numbers. $$\Delta_C = \frac{864 + 189}{EI} = \frac{1053}{EI} = \frac{1053}{3.0 \times 10^{4}} = 0.0351\ \text{m}$$ $$\boxed{\Delta_C = 35.1\ \text{mm downwards}}$$ The tip load supplies 28.8 mm of that and the partial UDL only 6.3 mm, because the distributed load sits close to the built-in end where the unit-load lever arm is small.