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16-Civ-A1 Elementary Structural Analysis · December 2013

Question 5 of 8: Indeterminate frame by moment distribution / slope–deflection

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

Notes on this paper

National Exams, December 2013 — 98-Civ-A1 Elementary Structural Analysis. Three-hour, closed-book examination (approved Sharp/Casio calculator only). Format: six questions constitute a complete paper — answer all of #1–#5 and any one of #6, #7 or #8. All eight questions are solved below for completeness.

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

Question 5: Indeterminate frame by moment distribution / slope–deflection (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/m 16 kN 4 kN 1 2 3 4 5 8 m 6 m 1 m 3 m
Frame 5: horizontal beam fixed at node 1; a vertical prop 2–5 (roller at 5) at x = 8 m; roller at node 3 (x = 14 m); 6 kN/m over 1–2, 16 kN at x = 11 m, 4 kN at the tip (node 4).

Given. Horizontal beam 1–2–3–4 with node 1 fixed, a vertical prop 2–5 (3 m long, roller at foot 5) at x = 8 m, and a roller at node 3 (x = 14 m); 1 m overhang to the free tip (node 4). Loads: 6 kN/m over span 1–2 (0–8 m), 16 kN at x = 11 m (mid of span 2–3), 4 kN at the tip. Two degrees of static indeterminacy.

Find. Member end moments, SFD/BMD ordinates, and reactions.

Approach. Solve the two rotational unknowns at the free joints (2 and 3) by slope–deflection, using fixed-end moments for the span loads. The prop 2–5 (roller foot) develops only axial force, so it contributes no bending stiffness — it simply provides a vertical reaction at node 2.

  1. Fixed-end moments. Span 1–2 ($w=6$, $L=8$): $\text{FEM} = wL^2/12 = 32$ kN·m. Span 2–3 ($P=16$ at mid, $L=6$): $\text{FEM} = PL/8 = 12$ kN·m. Overhang 3–4 applies a fixed $4(1)=4$ kN·m at node 3.
  2. Joint balance (moment distribution / slope–deflection). Solving the joint-rotation equations at nodes 2 and 3 gives the member end moments (hogging negative):
  3. Member 1–2: $M_{12}=-36$ kN·m at the fixed wall, $M_{21}=-24$ kN·m at node 2.
  4. Member 2–3: $M_{23}=-24$ kN·m at node 2, $M_{32}=-4$ kN·m at node 3 — joint 2 balances since $M_{21}+M_{23}=0$ (the prop carries no moment).
  5. Member 3–4: the statically determinate overhang delivers exactly $-4$ kN·m at node 3, balancing $M_{32}$.
  6. Shears / reactions. Span 1–2: $V=+25.5$ kN at the wall, zero at x = 4.25 m, $-22.5$ kN at node 2. Span 2–3: $+11.33$ then $-4.67$ kN across the 16 kN load. Overhang: $+4$ kN. Reactions: wall node 1 $\to V=25.5$ kN ↑, $M=36$ kN·m; prop foot node 5 $\to 33.83$ kN ↑; roller node 3 $\to 8.67$ kN ↑ (sum $=68$ kN $=6\!\times\!8+16+4$ ✓).
−36 +18 −24 +10 −4
BMD 5 (developed along the beam): hogging −36 at the wall, sagging +18 mid-span 1–2, hogging −24 at node 2, sagging +10 under the 16 kN load, hogging −4 at node 3.
LocationBending moment (kN·m)
Fixed wall (node 1)−36 (hogging) — member 1–2 minimum
Mid span 1–2 (x≈4.25 m)+18 (sagging) — maximum
Node 2−24 (hogging)
Under 16 kN (x = 11 m)+10 (sagging)
Node 3−4 (hogging)
Prop 2–50 (axial member, 33.8 kN compression)