Question 4 of 9: Deflection at an internal hinge by Castigliano's theorem
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2017 — 07-Str-A4 Advanced Structural Analysis. Three hours, CLOSED BOOK (an approved Sharp or Casio calculator is permitted). Answer BOTH Questions 1 and 2, ONLY TWO of Questions 3, 4 or 5, and ONLY TWO of Questions 6, 7, 8 or 9; six questions constitute a complete paper for 100 marks. Marks are printed in the left margin. All nine questions are worked below, because the complete set is the study resource.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (Pearson) — Ch. 8 influence lines, Ch. 9–10 the force (flexibility) method, Ch. 11 slope-deflection, Ch. 14–16 the stiffness method; A. Kassimali, Structural Analysis, 6th ed. (Cengage) — Ch. 7 deflections by work and energy (Castigliano), Ch. 13 slope-deflection, Ch. 16 sidesway frames; W. McGuire, R. H. Gallagher & R. D. Ziemian, Matrix Structural Analysis, 2nd ed., for the element stiffness matrix used as the independent check. Once the analysis is complete, member design follows CSA A23.3:19, CSA S16:19 and CSA O86:19, with load combinations from NBCC 2020.
Check: sign convention used throughout. End moments $M_{ij}$ are counter-clockwise positive on the member end — the convention that matches the six-degree-of-freedom planar frame element stiffness matrix, so every answer below can be checked against a direct-stiffness solution. Chord rotation is $\psi_{ij}=\left[(\mathbf{D}_{j}-\mathbf{D}_{i})\cdot\mathbf{e}_{2}\right]/L$, where $\mathbf{e}_{2}$ is the member axis turned $+90^\circ$. The fixed-end moment of a downward uniform load is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end and $-wL^{2}/12$ at the $j$ end; for a member released at its far end it becomes $+wL^{2}/8$. Ordinary sagging moments follow as $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$, and all diagrams are plotted sagging positive. Member shear at the $i$ end is $V_{i}=(M_{ij}+M_{ji})/L$ plus the equivalent nodal shear of any span load. Mixing this with Hibbeler's clockwise-positive convention produces clean-looking integers that are wrong, so the convention is stated once and used everywhere.
Question 4: Deflection at an internal hinge by Castigliano's theorem (16 marks)
Given. A built-in end at A, an internal hinge at B 4 m away, a roller at C a further 10 m along, a uniform load of 4 kN/m over A–B, a 36 kN point load 5 m from C, and $EI=51\,200$ kN·m$^{2}$ throughout.
Find. The vertical deflection of point B, the internal hinge.
The hinge at B splits the member into a simply supported span B–C and a loaded cantilever A–B; the span reaction rides on the cantilever tip.
Approach. Use the hinge to make the structure determinate, transfer the span reaction onto the cantilever tip as a point load, then apply Castigliano's second theorem with a dummy force at B.
Split the structure at the hinge. The moment at B is zero and the moment at the roller C is zero, so B–C is a simply supported span of 10 m carrying 36 kN at 5 m from C. Its left-hand reaction is$$R_{B}=\frac{36(5)}{10}=18\ \text{kN},$$and by Newton's third law that 18 kN presses down on the tip of the cantilever A–B. The span also carries a peak sagging moment of $18(5)=90$ kN·m under the load.
Set up the moment expression with a dummy force. Measure $x$ from B back towards A and apply a downward dummy force $Q$ at B. The cantilever moment is$$M(x)=-\left[\frac{wx^{2}}{2}+(R_{B}+Q)\,x\right],\qquad \frac{\partial M}{\partial Q}=-x .$$The span B–C contributes nothing, because its moments do not depend on $Q$ — the hinge cannot transmit the extra force back into the span.
Integrate Castigliano's second theorem. With $Q$ set to zero after differentiating,$$\Delta_{B}=\int_{0}^{L}\frac{M}{EI}\frac{\partial M}{\partial Q}\,dx=\frac{1}{EI}\int_{0}^{4}\left(2x^{3}+18x^{2}\right)dx=\frac{1}{EI}\Big[\tfrac{1}{2}x^{4}+6x^{3}\Big]_{0}^{4}.$$Evaluating the bracket gives $128+384=512$ kN·m$^{3}$, so$$\Delta_{B}=\frac{512}{51\,200}=\boxed{0.0100\ \text{m}=10.0\ \text{mm downwards}}.$$The two terms are recognisable as the standard cantilever results $wL^{4}/8EI=128/EI$ and $PL^{3}/3EI=384/EI$, which is the quickest independent check.
Complete the force picture. The cantilever also fixes the built-in end, where$$M_{A}=\frac{wL^{2}}{2}+R_{B}L=\frac{4(4)^{2}}{2}+18(4)=104\ \text{kN}\cdot\text{m}$$hogging, with a reaction of $4(4)+18=34$ kN up. A direct-stiffness model with the hinge represented by a single end release reproduces both the 10.0 mm deflection and the 104 kN·m fixing moment.
