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16-Civ-B1 Advanced Structural Analysis: May 2017

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

  1. Question 1 Statical indeterminacy and structural degrees of freedom
  2. Question 2 Schematic shear force and bending moment diagrams
  3. Question 3 Three-bar truss by the theorem of least work
  4. Question 4 Deflection at an internal hinge by Castigliano's theorem
  5. Question 5 Plane frame jacked at mid-span
  6. Question 6 Sway frame with load-carrying stubs, by slope-deflection
  7. Question 7 Fixed-end moment of a non-prismatic beam by the flexibility method
  8. Question 8 Trapezoidal sway frame with an inextensible strut
  9. Question 9 Deriving the stiffness matrix and load vector

Start with Question 1 →

Paper format. National Exams, May 2017 — 16-Civ-B1 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.