24-Bld-A1 Elementary Structural Analysis · December 2017
Question 8 of 8: Moment distribution for a symmetric multi-bay frame with two internal hinges (22 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2017 — 07-BLD-A1 Elementary Structural Analysis, 3 hours, closed book. Answer ALL of Questions 1–5; answer ONLY ONE of Questions 6, 7 or 8 (this solution set, per pipeline convention, answers all three optional questions).
Given. A continuous top beam (26 m, UDL 12 kN/m throughout, plus a 4.8 kN point load at mid-span x=13 m) carried on four 4 m columns: pin bases at x=0 and x=26, fixed bases at x=8 and x=18; internal hinges in the beam at x=10 and x=16 (a 6 m "drop-in" span between them). The geometry and loading are symmetric about x=13 m.
Find. Member end moments and the shear/moment diagrams, using symmetry to halve the work.
Symmetric frame: pin/fixed/fixed/pin bases, hinges at x=10, 16 m, UDL 12 kN/m + 4.8 kN at mid-span.
Approach. The two hinges make this a classic three-part "Gerber" system: the central 6 m span is simply supported at the hinges and is fully determinate on its own; solving it first gives the point loads the outer sub-frames must carry at their hinge tips. Because the whole structure and its loading are symmetric about x=13 m, the two outer sub-frames (0–10 m and 16–26 m) are exact mirror images, so only the left one need be analysed.
Central drop-in span (10 m to 16 m, simply supported at the hinges). By symmetry each hinge carries half the span's own load: $$V_{hinge}=\tfrac12\left[12(6)+4.8\right]=\tfrac12(76.8)=\boxed{38.4\text{ kN}}$$ transmitted as a downward point load onto each outer sub-frame's hinge tip.
Left sub-frame (pin at x=0/column up to the beam at x=0–8, fixed column at x=8, 2 m beam extension to the hinge at x=10 carrying the 38.4 kN tip reaction, plus UDL 12 kN/m over the full 0–10 m). This is a portal frame with one pinned and one fixed base — indeterminate to the 2nd degree; solved here by an equivalent slope-deflection/moment-distribution solve (member stiffnesses equal, tabulated numerically and cross-checked with a full plane-frame stiffness analysis):
Location (left sub-frame)
Value
Reaction at pin base (x=0)
F_x=5.71 kN, F_y=38.21 kN
Reaction at fixed base (x=8)
F_x=−5.71 kN, F_y=120.19 kN, M=22.48 kN·m
Moment at top of pin column (x=0)
22.84 kN·m
Moment in beam at joint (x=8), beam side
101.2 kN·m
Moment in beam at joint (x=8), column side
−0.4 kN·m (column carries almost none)
Zero-shear / peak hogging in span 0–8
−38.0 kN·m at x=3.18 m
Moment at hinge (x=10)
0 (free/hinge condition)
Right sub-frame (16–26 m): by symmetry, identical magnitudes mirrored about x=13 m — reaction at the pin base (x=26) is F_x=−5.71 kN, F_y=38.21 kN; at the fixed base (x=18), F_x=5.71 kN, F_y=120.19 kN, M=22.48 kN·m (opposite sense, mirrored).
