16-Civ-A1 Elementary Structural Analysis · December 2017
Question 8 of 8: Moment distribution of a symmetric Gerber frame
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Civ-A1 Elementary Structural Analysis, December 2017, 3 hours, closed book (approved Sharp or Casio calculator only). A complete paper is six questions: candidates answer all of Q1–Q5 and exactly one of Q6, Q7 or Q8. All eight questions are worked below — Q6, Q7 and Q8 each in full — so the set is a complete study resource. Marks: Q1 [6], Q2–Q5 [18 each], Q6–Q8 [22 each].
Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy & stability (Ch. 2), method of joints/sections (Ch. 3), shear & moment diagrams (Ch. 4), deflections by virtual work (Ch. 8–9), influence lines (Ch. 6), slope–deflection and moment distribution (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — compound (Gerber) beams, internal hinges and condition equations, indeterminate frames. Sign convention: sagging bending moment positive (tension on the underside); truss forces tension (T) positive, compression (C) negative.
Reading the exam figures. Every support, member and load below was read directly from the printed figures. The Q4(b) truss carries a stray “3 m” label that is the perpendicular spacing between the two parallel chords (confirmed below), and the Q8 frame carries a stray “4.8 m” dimension that is not a load — it is the offset from the beam down to the lower dimension line (the columns are 4 m, stated on both outer columns).
Question 8 — Moment distribution of a symmetric Gerber frame [22]
Continuous top beam (26 m) on four 4 m columns: pin (1), fixed (4), fixed (8), pin (7). UDL 12 kN/m over the whole beam. Two internal hinges split the central span into 2 m cantilevers + a 6 m suspended span.
Given. Symmetric frame about mid-span. Top beam spans $8+10+8$ m; the central $10$ m span carries two internal hinges $2$ m from each interior column, leaving a $6$ m suspended span between them. Columns $4$ m: outer bases pinned (1, 7), inner bases fixed (4, 8). UDL $12$ kN/m over the entire beam.
Find. Member end moments and the $V$, $M$ diagrams.
Approach. The two hinges make the $6$ m centre a simply supported (Gerber) span that delivers point loads to the cantilever tips; the rest is an ordinary two-bay frame solved by moment distribution, using symmetry to halve the work.
Suspended span. $6$ m under $12$ kN/m: end reactions $\tfrac{12(6)}{2}=36$ kN delivered as downward loads at each hinge; its own $M_{\max}=\dfrac{wL^2}{8}=\dfrac{12(6)^2}{8}=\boxed{54\text{ kN}\cdot\text{m}}$ (sag).
Cantilever root (interior column, joint 3). The $2$ m cantilever carries its own UDL plus the $36$ kN hinge reaction at its tip: $M=36(2)+12(2)\tfrac{2}{2}=72+24=\boxed{-96\text{ kN}\cdot\text{m}}$ (hogging).
Outer span 2–3 by moment distribution. Fixed-end moments from the $8$ m UDL ($\pm64$) plus the $-96$ carry-in from the cantilever are distributed among the beam and the two columns at joint 3 (pinned base column 1–2 has stiffness $3EI/4$; fixed base column 3–4 uses $4EI/4$). Balancing gives $M_{32}=-82.8$ kN·m and $M_{23}=-36.4$ kN·m.
Span extremes. Beam 2–3 shear $+42.2\to-53.8$ kN, so max sag $+37.8$ kN·m; the mirror span 5–6 is identical. Column moments: pinned column top $-36.4$; fixed inner column $-13.2$ (top) / $+6.6$ (base).
Reactions. Vertical: outer (pinned) columns $42.2$ kN, inner (fixed) columns $113.8$ kN — total $2(42.2)+2(113.8)=312=12(26)$ &checkmark. Column shears give the small horizontal reactions ($9.1$ kN at the pins, $5.0$ kN at the fixed bases).
Bending moment (top beam). Hogging $-96$ at each interior column (cantilever root), $-82.8$ in the outer beam at that joint, $-36.4$ at the outer columns; sag $+37.8$ in the outer spans and $+54$ in the suspended centre span; $M=0$ at both hinges.
Shear force (top beam): outer span $+42.2\to-53.8$ kN; suspended span $\pm36$ kN; antisymmetric about mid-span.
Location
Moment (kN·m)
Interior column joints 3, 5 (cantilever root)
$-96$ (hogging, governs)
Outer beam at 3, 5
$-82.8$
Outer spans 2–3, 5–6 max sag
$+37.8$
Suspended span mid
$+54$ (hinges $M=0$)
Outer columns (pin base) top
$-36.4$
Reactions: outer / inner columns
$42.2$ / $113.8$ kN (vertical)
Check. Modelling the two hinges as end-moment releases (one release per hinge, so no mechanism), the finite-element solution gives the suspended-span mid moment $54$, cantilever root $-96$, outer-joint $-82.76$ and reaction sum $312$ kN — matching the hand moment-distribution values.
