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16-Civ-A1 Elementary Structural Analysis · May 2018

Question 7 of 8: Frame with relative $EI$ (slope-deflection / moment distribution)

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

Notes on this paper

Paper: National Exams — May 2018, 16-Civ-A1 Elementary Structural Analysis (closed book, 3 h; Casio/Sharp calculator permitted). Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy (Ch. 2), method of joints & sections (Ch. 3), shear & moment diagrams (Ch. 4), influence lines (Ch. 6), virtual-work deflections (Ch. 8–9), moment distribution / slope–deflection (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, indeterminate frames.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside), hogging negative; member axial force tension positive (T), compression negative (C).

These A1 papers are defined entirely by their figures, so every structure is redrawn to scale below.

Question 7: Frame with relative $EI$ (slope-deflection / moment distribution) (22 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Inclined member $1\!-\!2$ ($1.3EI$, rise $2.5\text{ m}$ over $6\text{ m}$) fixed at 1; a vertical strut $2\!-\!5$ ($EI$, $3\text{ m}$) pinned at 5; a horizontal beam $2\!-\!3\!-\!4$ ($1.4EI$) with a roller at 3 ($8\text{ m}$ from 2) and a free tip 4 ($3\text{ m}$ beyond); UDL $10.8\text{ kN/m}$ over span $2\!-\!3$ and $14.3\text{ kN}\downarrow$ at tip 4. Find. reactions, SFD, BMD.

10.8 kN/m14.3 kN123451.3EI1.4EIEI6 m8 m3 m
Q7 — frame with three different relative stiffnesses.
  1. Statically-known tip moment. The $3\text{ m}$ overhang $3\!-\!4$ carries only the tip load, so the moment over the roller at 3 is fixed by statics: $M_{@3}=14.3(3)=\boxed{42.9\text{ kN}\cdot\text{m}}$ (hogging).
  2. Slope-deflection / distribution. With member stiffnesses $1.3EI/6.5,\;EI/3,\;1.4EI/8$ meeting at joint 2, solving the joint rotations (the far ends at 1 fixed, 5 and 3 pinned) gives the joint-2 moments $$\boxed{M_{2,\text{beam}}=50.3,\quad M_{2,\text{strut}}=27.9,\quad M_{2,\text{incline}}=22.4\text{ kN}\cdot\text{m}}$$ which balance at the joint.
  3. Reactions. $$\boxed{V_5=53.6\text{ kN},\;V_3=56.6\text{ kN},\;V_1=-9.5\text{ kN (down)}}\quad(\Sigma V=100.7\text{ kN}=10.8\times8+14.3);$$ fixed base 1 also carries $H_1=9.3\text{ kN}$ and $M_1=11.2\text{ kN}\cdot\text{m}$.
  4. Diagram extremes. Span $2\!-\!3$: hogging $50.3$ at 2, hogging $42.9$ at 3, sagging $\approx +40\text{ kN}\cdot\text{m}$ near mid-span (shear zero at $\approx4.1\text{ m}$); the strut $2\!-\!5$ carries a linear moment $27.9\to0$; the inclined member $1\!-\!2$ runs $11.2\to22.4\text{ kN}\cdot\text{m}$.

Check: the UDL is read as acting over the main beam span $2\!-\!3$; the $3\text{ m}$ tip carries only the $14.3\text{ kN}$ point load.