Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examinations, December 2015 — 98-Civ-B1 Advanced Structural Analysis. Three hours, closed book (an approved Sharp or Casio calculator is the only aid). Nine questions: Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers two of Questions 3, 4, 5 (18 marks each) and two of Questions 6, 7, 8, 9 (22 marks each), so six questions constitute a complete 100-mark paper. All nine questions are solved here, because this set is a study resource rather than an examination script.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection, moment distribution, influence lines, Castigliano); A. Kassimali, Structural Analysis, 6th ed. (force and displacement methods, degrees of freedom); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (flexibility and stiffness formulations, lack of fit, support movement); W. Weaver Jr. & J. M. Gere, Matrix Analysis of Framed Structures, 3rd ed. (assembly of [K] and {P}); J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods, 4th ed. Canadian design context: CSA S16, CSA A23.3 and the National Building Code of Canada supply the load and resistance rules that these analyses feed, although this paper is purely an analysis paper.
Check: sign convention used throughout. Member end moments \(M_{ij}\) are counter-clockwise positive, which is the convention that matches the standard six-degree-of-freedom stiffness matrix and the vector \(\{\theta_2,\theta_3,\delta\}\) the paper itself defines (Question 9 states “positive counter clockwise”). With that choice the fixed-end moment of a downward uniformly distributed load is \(\mathrm{FEM}_{ij}=+wL^{2}/12\) at the \(i\) end, the chord rotation is \(\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_\perp]/L\) with \(\mathbf{e}_\perp\) the member axis turned \(+90^{\circ}\), and ordinary sagging moments are recovered as \(M_{\text{sag}}(i)=-M_{ij}\), \(M_{\text{sag}}(j)=+M_{ji}\). Mixing this with Hibbeler’s clockwise-positive convention produces clean-looking integers that are wrong; every diagram below is plotted in the sagging convention.
Check: printed unit in Question 6. The paper prints “EI = 3.2 × 104 kN.mm2”. Taken literally that is a flexural rigidity some ten orders of magnitude too small for an 18 m beam and would make the 36 mm settlement effect vanish. The intended unit is \(\text{kN}\cdot\text{m}^2\): with \(EI=3.2\times10^{4}\ \text{kN}\cdot\text{m}^{2}\) the settlement term \(3EI\Delta/L^{2}=54\ \text{kN}\cdot\text{m}\) is the same order as the load term \(wL^{2}/8=48\ \text{kN}\cdot\text{m}\), which is plainly what the question is testing. The solution below uses \(\text{kN}\cdot\text{m}^{2}\) and flags the typographical error.