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16-Civ-A1 Elementary Structural Analysis · May 2014

Question 5 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, 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).

The paper directs candidates to answer Q1–Q4, then one of Q5/Q6 and one of Q7/Q8. For completeness all eight questions are worked here.

Question 5: 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.

24 kN/m12 (hinge)34567821695: continuous beam 1–5 (4EI) on two EI columns fixed at 6,7; internal hinge at node 2; UDL 24 kN/m over the whole beam.

Given. Continuous beam, nodes $1(x{=}0,\text{pin})$–$2(8,\text{hinge})$–$3(10)$–$4(26)$–$5(35,\text{roller})$, all spans $4EI$; columns $3\!\to\!6$ and $4\!\to\!7$, each 4 m long, $EI$, fixed at the base. UDL $w=24$ kN/m over the entire beam. Structure is 5° statically indeterminate.

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

Approach. The internal hinge at node 2 makes span 1–2 a simply-supported carry-over: its right-end shear ($wL/2=96$ kN) and the 2 m stub 2–3 deliver a known moment into joint 3, leaving joints 3 and 4 (each a beam–beam–column node) as the rotating unknowns for moment distribution. Distribution factors use $k_{\text{beam}}=4EI_b/L$ and $k_{\text{col}}=4EI_c/L$ (far ends fixed at 6,7); with no sidesway (pin at 1, inextensible columns) the balance/carry-over converges to the member-end moments below.

  1. Determinate carry-in from the hinge. Span 1–2 (pin–hinge, 8 m): simply supported under the UDL, mid-span sagging $wL^2/8=\boxed{+192\ \text{kN}\cdot\text{m}}$, hinge shear $96$ kN. The 2 m stub 2–3 then delivers $M_3 = -[96(2)+24(2)(1)] = \boxed{-240\ \text{kN}\cdot\text{m}}$ to joint 3 (from the left).
  2. Distribute at joints 3 and 4. Balancing the unbalanced moments through the two columns and the long 16 m span gives the member-end moments (beam faces): $M_{3,\text{right}}=-402$, $M_{4,\text{left}}=-488$, $M_{4,\text{right}}=-383$ kN·m; column tops $M_{3\text{-}6}=+162$, $M_{4\text{-}7}=-105$ kN·m; column bases (fixed) $M_6=-81$, $M_7=+53$ kN·m.
  3. Span moments. On 3–4 the sagging peak is $\boxed{+323\ \text{kN}\cdot\text{m}}$ at $x\approx17.8$ m; on 4–5 the sagging peak is $+89$ kN·m near $x\approx32$ m; the largest hogging is $\boxed{-488\ \text{kN}\cdot\text{m}}$ over joint 4.
  4. Reactions. $R_1=96\uparrow$, $R_5=65.4\uparrow$ (vertical); pin horizontal $H_1=21.5$ kN. Column feet: node 6 $V=331\uparrow,H=61,M=-81$; node 7 $V=348\uparrow,H=-39,M=+53$. ($\sum V=840=24\times35$ ✓, $\sum H=0$ &checkmark.)
SFD beam 1-5 (kN)+96-144+187-197+151-65 BMD beam 1-5 (kN·m) (steps at 3,4 = column moments)+192-240+323-488+89
LocationBending moment (kN·m)
Span 1–2 mid$+192$ (sagging)
Node 2 (hinge)$0$
Node 3 — beam left / right$-240$ / $-402$
Span 3–4 peak$+323$
Node 4 — beam left / right$-488$ / $-383$
Span 4–5 peak$+89$
Column 3–6 top / base$+162$ / $-81$
Column 4–7 top / base$-105$ / $+53$

Check (model): node 2 is read as a true internal hinge (moment = 0), columns as rigidly framed into the beam and fixed at 6,7, and node 1/5 as pin/roller. These readings make the frame 5° indeterminate; the member-end moments above satisfy joint balance at 3 and 4 and the fixed-base conditions exactly.