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07-Str-A4: December 2017

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

  1. Question 1 Minimum number of structural degrees of freedom (8 marks)
  2. Question 2 Schematic shear force and bending moment diagrams (12 marks)
  3. Question 3 Castigliano’s theorem — horizontal deflection at joint 3 (16 marks)
  4. Question 4 Least work — forces in a three-bar truss (16 marks)
  5. Question 5 Slope-deflection for a frame jacked at joint 3 (16 marks)
  6. Question 6 Symmetric frame with internal hinges, by slope-deflection (24 marks)
  7. Question 7 Flexibility method — fixed-end moment of a non-prismatic beam (24 marks)
  8. Question 8 Symmetric frame with a Gerber top member (24 marks)
  9. Question 9 Derive the stiffness matrix and load vector (24 marks)

Start with Question 1 →

Paper format. National Examinations, December 2017 — 07-Str-A4 Advanced Structural Analysis. Three hours, CLOSED BOOK (an approved Sharp or Casio calculator is the only aid). Nine questions: answer BOTH #1 and #2, then ONLY TWO of #3, #4 or #5, and ONLY TWO of #6, #7, #8 or #9 — six questions constitute a complete paper. Marks are printed in the left margin: 8, 12, 16, 16, 16, 24, 24, 24, 24. All nine questions are worked below, because the complete set is the study resource rather than a sitting.

Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (Pearson) — Ch. 4 shear and moment diagrams, 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 and least work), Ch. 13 slope-deflection, Ch. 16 sidesway frames; W. McGuire, R. H. Gallagher and R. D. Ziemian, Matrix Structural Analysis, 2nd ed., for the planar frame element stiffness matrix used as the independent check on every answer below. Member sizing that would follow such an analysis is governed in Canada by CSA A23.3:19, CSA S16:19 and CSA O86:19, with load combinations from NBCC 2020.

Check: sign convention, stated once and 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, so every result below can be reproduced by a direct-stiffness solution. The chord rotation is $\psi_{ij}=\left[(\mathbf{D}_{j}-\mathbf{D}_{i})\cdot\mathbf{n}\right]/L$ with $\mathbf{n}$ the member axis turned $+90^\circ$; the fixed-end moment of a downward UDL is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end; and the ordinary sagging moment is $M_{\text{sag}}(i)=-M_{ij}$, $M_{\text{sag}}(j)=+M_{ji}$. Mixing this with the clockwise-positive convention produces clean-looking integers that are wrong.