Question 1 of 8: Stability and Degree of Static Indeterminacy

Question 1: Stability and Degree of Static Indeterminacy (6 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.

Given. Six plane structures. (a) A beam carrying a uniform load over its whole length, free at the left, on two rollers, with two internal hinges (the arrow labels them “typical hinge”) and built in at the right-hand wall. (b) A two-storey, single-bay rigid frame, both floors loaded, pinned at the left base and fixed at the right base. (c) One continuously rigid bent member: a loaded horizontal beam on two rollers, turning up at its right end and back to the left along an upper limb that is pinned to the ceiling. (d) A loaded lower beam, pinned at its left end and on a roller at its right end, rigidly continuous with a column that rises to an internal hinge; above the hinge a loaded upper beam runs to a second column fixed at its base. (e) A pin-jointed truss of 13 members and 8 joints on a pin and a roller. (f) A pin-jointed truss: a square panel carrying both diagonals with a triangle built on top, on a pin and a roller, loaded horizontally at the apex.

Find. For each structure, its classification — unstable, statically determinate, or statically indeterminate — with the degree of indeterminacy where that applies.

[Figure not reproduced: Question 1 — the six structures, redrawn with the support symbols and internal hinges read off the examination figure. See the official exam paper.]

Approach. Count first, then look: for beam- and frame-type structures use $i = 3m + r - 3j - c$ (for a single unbranched member this collapses to $i = r - 3 - c$), for pin-jointed trusses use $i = m + r - 2j$, and then confirm that the constraints are actually arranged so as to prevent every rigid-body motion, because a favourable count never by itself proves stability.

1. Structure (a) — Gerber beam, two hinges. The beam is a single line of members, so only the reactions and the release conditions matter. Two rollers and one fixed end give $r = 1 + 1 + 3 = 5$, and the two internal hinges supply $c = 2$ condition equations, so $$i = r - 3 - c = 5 - 3 - 2 = \boxed{0}$$ The arrangement is the classical suspended-span layout: the left portion is a beam with an overhang on two rollers, the middle portion between the hinges is a simply supported drop-in span, and the right portion is a cantilever off the wall. Horizontal restraint comes from the fixed end and travels through the hinges, which transmit axial force. Statically determinate.
2. Structure (b) — two-storey single-bay rigid frame. Split each column at the intermediate floor: $m = 6$ members (four column segments and two beams) joining $j = 6$ joints, with $r = 2 + 3 = 5$ from the pinned and the fixed base and no internal releases. Then $$i = 3m + r - 3j - c = 3(6) + 5 - 3(6) - 0 = \boxed{5}$$ The same number follows from the ring count: two closed rings at three redundants each is six, less one for the pin at the left base. Statically indeterminate to the fifth degree.
3. Structure (c) — single bent member on three restraints. Every corner is rigid and there is no release anywhere, so the whole thing is one member. The two rollers under the lower beam give one reaction component each and the ceiling connection is a pin, worth two, hence $r = 1 + 1 + 2 = 4$ and $$i = r - 3 = 4 - 3 = \boxed{1}$$ The three reaction lines are neither concurrent nor all parallel — two vertical rollers plus a pin above — so the extra restraint is a genuine redundancy and not a substitute for a missing one. Statically indeterminate to the first degree.
4. Structure (d) — two levels joined by a hinge. Take $m = 4$ (lower beam, lower column, upper beam, upper column) meeting at $j = 5$ joints, with $r = 2 + 1 + 3 = 6$ from the pin, the roller and the fixed base, and $c = 1$ for the hinge at the top of the lower column: $$i = 3m + r - 3j - c = 3(4) + 6 - 3(5) - 1 = \boxed{2}$$ The reading that matters here is that the lower beam is rigidly continuous with the column above the roller — that L-shaped piece already carries three reactions of its own and receives two force components through the hinge, which is where the two redundants sit. Statically indeterminate to the second degree.
5. Structure (e) — pin-jointed truss, 13 members. Counting off the drawing: four top-chord members, two bottom-chord members, three verticals and four diagonals give $m = 13$; there are $j = 8$ joints and $r = 3$ from the pin and the roller, so $$i = m + r - 2j = 13 + 3 - 2(8) = \boxed{0}$$ Stability follows by construction: the triangle at the left support is rigid, two further joints are added two members at a time, and the remaining triangle at the right is tied back by three non-concurrent, non-parallel links. Statically determinate.
6. Structure (f) — braced square with an apex triangle. The panel carries both diagonals, and the note tells us they are not connected where they cross, so both are real members: $m = 8$ (four sides of the square, two diagonals, two rafters), $j = 5$ and $r = 3$, hence $$i = m + r - 2j = 8 + 3 - 2(5) = \boxed{1}$$ The redundancy is internal — the panel is braced twice over — while the external reactions are exactly sufficient. Statically indeterminate to the first degree.