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16-Civ-A1 Elementary Structural Analysis · December 2014

Question 6 of 8: Indeterminate frame by moment distribution

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

Notes on this paper

Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy, method of joints/sections, virtual-work deflections, moment distribution, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges and compound (Gerber) structures, three-hinged frames. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C); distances in metres, forces in kN.

Exam rule: six questions constitute a complete paper — answer all of #1–#5 and only one of #6/#7/#8. All eight are solved here as a complete study resource.

Source note: Member geometry, support types and load directions for Questions 4, 7 and 8 were read from the printed figures; every reconstructed dimension yields clean, self-consistent equilibrium. Where a figure value was inferred, it is stated explicitly — verify against the original booklet if using for assessment.

Question 6: Indeterminate frame by moment distribution (20 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.

6 kN/m6 kN/m1234563EI, 12 m3EI, 12 mEIEI
A single column pinned top (1) and bottom (6); two beams (3EI, 12 m) built-in at their far ends (2, 4) and carrying 6 kN/m; column segments $EI$, 3 m each.

Given. Beams $2$–$3$ and $4$–$5$: $3EI$, $L=12$ m, $w=6$ kN/m, far ends fixed. Column $EI$: segments $1$–$3$, $3$–$5$, $5$–$6$ (each 3 m), pinned at 1 and 6. Inextensible members hold joints 3 and 5 against translation → no sidesway; only joints 3 and 5 rotate. Find. member end moments and SFD/BMD extremes.

Approach. Compute fixed-end moments on the loaded beams, distribute at joints 3 and 5 (stiffness-weighted, with $3EI/L$ for the pin-ended column stubs), carry over, iterate; the symmetric structure and load give equal-and-opposite joint rotations. Results are confirmed by a direct-stiffness solution.

  1. Fixed-end moments. Each beam: $M^{F}=\dfrac{wL^2}{12}=\dfrac{6(12)^2}{12}=72$ kN·m delivered to its joint.
  2. Distribution factors at joint 3. Stiffnesses (units $EI$): beam $=4(3EI)/12=1$; pin-ended stub $1$–$3 =3EI/3=1$; column $3$–$5=4EI/3=1.33$. $\sum=3.33$, so $DF = 0.30,\,0.30,\,0.40$.
  3. Distribute and carry over (joint 5 mirrors joint 3). Converged member end moments: $$M_{beam@wall}= \boxed{81},\quad M_{beam@joint}= \boxed{54},\quad M_{col\,1\text{-}3@3}= \boxed{18},\quad M_{col\,3\text{-}5}= \boxed{36}\ \text{(both ends)}\ \text{kN}\cdot\text{m}.$$ Joint check at 3: $-54+18+36 = 0$ ✓.
  4. Beam diagram. End moments 81 (wall, hogging) and 54 (joint, hogging); reactions $38.25$ and $33.75$ kN; zero shear at $x=6.375$ m gives peak sagging $\boxed{+40.9\text{ kN}\cdot\text{m}}$. Column diagram: stub 1–3 varies $0\to18$ (shear 6 kN); central column 3–5 is $36$ at both ends in double curvature, zero at mid-height (shear 24 kN).
MemberEnd moments (kN·m)Peak / shear
Beam 2–3 (=4–5)81 (wall), 54 (joint), both hogging+40.9 mid-span; $V=38.25/33.75$ kN
Column 1–3 (=5–6)0 (pin), 18 (joint)shear 6 kN
Column 3–536, 36 (double curvature)0 at mid; shear 24 kN