Question 4 of 8: Horizontal Deflection of a Hinged Portal by Castigliano’s Theorem

Question 4: Horizontal Deflection of a Hinged Portal by Castigliano’s Theorem (21 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.

Find. The horizontal displacement of joint 2, in magnitude and direction.

Approach. Confirm the frame is determinate, apply a dummy horizontal force \(Q\) at joint 2, write \(M\) and \(\partial M/\partial Q\) for every member, then evaluate \(\Delta = \int M\,(\partial M/\partial Q)\,\mathrm{d}s/EI\) at \(Q = 0\).

1. Check determinacy. Two pins give four reaction components against three equations, so the frame would be one degree indeterminate; the internal hinge at joint 2 supplies exactly one extra condition, so$$r_{\text{ext}} - 3 - (\text{releases}) = 4 - 3 - 1 = \boxed{0}$$and Castigliano’s theorem may be applied directly to a determinate structure.
2. Apply the dummy force and solve the statics. Let \(Q\) act horizontally at joint 2, positive to the right. Taking moments about the hinge for the free body consisting of column 1–2 alone gives \(H_1 h = 0\), so$$H_1 = 0,\qquad H_4 = -Q$$Column 1–2 is thus a two-force member: whatever else happens, it carries axial force only. Global moment equilibrium about joint 1 then gives$$V_4 = \frac{wL}{2} + \frac{hQ}{L} = 9 + Q,\qquad V_1 = \frac{wL}{2} - \frac{hQ}{L} = 9 - Q$$in kilonewtons for the given dimensions.
3. Write the internal moments member by member. Measuring \(s\) up from each base and \(x\) from joint 2 along the beam,$$M_{1\text{-}2}(s) = H_1 s = 0,\qquad M_{3\text{-}4}(s) = -Q\,s,\qquad M_{2\text{-}3}(x) = V_1 x - \frac{w x^{2}}{2}$$The hinge at joint 2 guarantees \(M = 0\) there, and \(Q\) applied at the hinge is purely axial to the beam, so it enters the beam moment only through \(V_1\).
4. Differentiate with respect to the dummy force.$$\frac{\partial M_{1\text{-}2}}{\partial Q} = 0,\qquad \frac{\partial M_{3\text{-}4}}{\partial Q} = -s,\qquad \frac{\partial M_{2\text{-}3}}{\partial Q} = -\frac{h}{L}\,x$$Setting \(Q = 0\) makes the column moments vanish identically, so the only member that contributes to the integral is the beam.
5. Integrate. With \(Q = 0\), \(V_1 = wL/2 = 9\) kN,$$\Delta_{H2} = \frac{1}{EI}\int_0^{L}\left(\frac{wL}{2}x - \frac{wx^{2}}{2}\right)\left(-\frac{h}{L}x\right)\mathrm{d}x = -\frac{w\,h\,L^{3}}{24EI}$$a closed form worth remembering. Substituting \(w = 6\) kN/m, \(h = L = 3\) m,$$\Delta_{H2} = -\frac{6(3)(3)^{3}}{24\,EI} = -\frac{20.25}{EI} = -\frac{20.25}{2.5\times 10^{4}} = \boxed{-8.10\times10^{-4}\ \text{m} = 0.81\ \text{mm to the left}}$$The negative sign means the movement opposes the assumed direction of \(Q\).
6. Confirm the result kinematically. Because both columns are moment-free, neither bends; each simply rotates rigidly about its pin. The beam is effectively simply supported, so its end rotation at the rigid corner 3 is$$\theta_3 = \frac{wL^{3}}{24EI} = \frac{6(27)}{24EI} = \frac{6.75}{EI}$$counter-clockwise. Column 3–4 must follow that rotation, and its head therefore moves \(h\theta_3 = 3(6.75)/EI = 20.25/EI\) to the left. The inextensible beam carries joint 2 with it, giving the same 0.81 mm — an independent derivation that uses no energy at all.
7. Round out the internal actions. With \(H_1 = H_4 = 0\) and \(V_1 = V_4 = 9\) kN, both columns carry 9 kN of axial compression and nothing else, and the beam is a simply supported span with$$M_{\max} = \frac{wL^{2}}{8} = \frac{6(9)}{8} = 6.75\ \text{kN}\cdot\text{m}$$at midspan. The frame is doing almost nothing as a frame: the hinge at joint 2 has switched off the portal action entirely.