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16-Civ-B1 Advanced Structural Analysis: May 2015

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

  1. Question 1 Schematic shear force and bending moment diagrams
  2. Question 2 Influence lines for a pin-jointed truss
  3. Question 3 Fixed-end moments of a non-prismatic beam by the flexibility method
  4. Question 4 Slope-deflection analysis of a frame with an imposed joint displacement
  5. Question 5 Two-beam structure by Castigliano’s least-work theorem
  6. Question 6 Continuous beam with a settling support
  7. Question 7 Frame with two columns and a loaded cantilever
  8. Question 8 Slope-deflection analysis of a frame on an inclined roller
  9. Question 9 Equilibrium equations and the stiffness matrix of a sway frame

Start with Question 1 →

Paper format. 98-Civ-B1 Advanced Structural Analysis, National Examinations, May 2015 — three hours, closed book (an approved Sharp or Casio calculator is the only aid). Questions 1 and 2 are compulsory; the candidate then answers two of Questions 3, 4, 5, one of Questions 6, 7 and one of Questions 8, 9, so six questions totalling 100 marks constitute a complete paper. Marks are printed in the left margin. All nine questions are solved here, because the set is a study resource rather than an examination script.

Reference texts. R. C. Hibbeler, Structural Analysis (Pearson) — force/flexibility method, slope-deflection, moment distribution and influence lines; A. Kassimali, Structural Analysis (Cengage) — matrix stiffness formulation and support-settlement effects; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods (Wiley); CSA S16:19 Design of Steel Structures and CSA S6:19 Canadian Highway Bridge Design Code for the Canadian design setting in which these analyses are used.

Sign convention used throughout. Bending moments are reported as sagging positive and are plotted on the tension face of each member. Slope-deflection end moments follow the counter-clockwise-positive convention stated in Question 9: $$M_{ij}=\frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi_{ij}\right)+M^{F}_{ij},\qquad \psi_{ij}=\frac{(\mathbf{D}_j-\mathbf{D}_i)\cdot \mathbf{e}_2}{L}$$ where $\mathbf{e}_2$ is the member axis rotated 90 degrees counter-clockwise, and the fixed-end moments of a downward uniform load are $M^{F}_{ij}=+wL^2/12$, $M^{F}_{ji}=-wL^2/12$. The sagging moment at the ends of a member is then $M(i)=-M_{ij}$ and $M(j)=+M_{ji}$.