Question 1 of 8: Determinacy and stability (6 marks)

Question 1 — Determinacy and stability (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 planar structures. In (a)–(d) the members carry bending (beam/frame action); in (e)–(f) all members are two-force truss bars and crossing diagonals are not connected. Open circles denote internal hinges; a plain triangle on hatching is a pin, a triangle over two wheels is a roller, and hatching alone is a built-in (fixed) support.

Find. A classification — unstable, determinate, or indeterminate to a stated degree — for each of the six structures.

Approach. Count released and restrained quantities: for beam and frame assemblies use the degree of static indeterminacy $\mathrm{DSI}=3m+r-3j-c$, and for pin-jointed trusses use $\mathrm{DSI}=m+r-2n$, then confirm by inspection that the restraints are arranged so the structure is actually stable (a non-negative count alone does not prove stability).

Here $m$ is the number of members, $j$ (or $n$) the number of joints, $r$ the number of independent support reaction components, and $c$ the number of equations of condition released by internal hinges. At a hinge joining $k$ members, $c=k-1$.

1. (a) Continuous beam with two internal hinges. The supports are one pin and three rollers, so $r = 2+1+1+1 = 5$, and the two hinges supply $c = 2$ equations of condition. For a single beam chain the count reduces to $$\mathrm{DSI} = r - 3 - c = 5 - 3 - 2 = \boxed{0}$$ The beam is statically determinate. It is also stable: the pin anchors the assembly horizontally, and each hinge-bounded segment is propped by at least one support, so no segment can move as a mechanism.

1. (b) Two-storey single-bay frame. Members: four column lengths (two per side, split at the lower-beam level) plus the two beams, so $m=6$; joints: two bases, two lower-beam ends and two top corners, so $j=6$. Both bases are fixed, giving $r=3+3=6$, and the lower beam is hinged at each end, giving $c=2$. Then $$\mathrm{DSI} = 3(6) + 6 - 3(6) - 2 = 18 + 6 - 18 - 2 = \boxed{4}$$ Indeterminate to the 4th degree. The same answer follows from the closed-loop rule: two closed loops give $2\times3=6$ redundants, less the two hinge releases.

1. (c) Propped beam on two fixed-base columns. Members: three beam lengths (between the two ends and the two column heads) plus two columns, so $m=5$; joints: two beam ends, two column heads and two bases, so $j=6$. The end supports are rollers ($1+1$) and both column bases are fixed ($3+3$), so $r=8$; two hinges give $c=2$. Then $$\mathrm{DSI} = 3(5) + 8 - 3(6) - 2 = 15 + 8 - 18 - 2 = \boxed{3}$$ Indeterminate to the 3rd degree.

1. (d) Hinged beam on two A-frames. Members: four beam lengths (the beam is divided by the four leg connections and the central hinge) plus four inclined legs, so $m=8$; joints: four leg-to-beam connections, the central hinge and the two apexes, so $j=7$. Each apex is a pin support, so $r=2+2=4$. Releases: the beam hinge contributes 1, and at each apex two legs meet on a common pin, contributing $2-1=1$ each, so $c=3$. Then $$\mathrm{DSI} = 3(8) + 4 - 3(7) - 3 = 24 + 4 - 21 - 3 = \boxed{4}$$ Indeterminate to the 4th degree. Physically, each A-frame closes a triangle with the beam segment above it, and the two pinned apexes provide one more horizontal restraint than statics needs.

1. (e) Triangular truss with crossing diagonals. Counting bars: two rafter lengths on each side (4), three bottom-chord panels (3), two verticals (2), the horizontal tie between the two mid-height joints (1) and the two crossing diagonals (2) give $m=12$ over $n=7$ joints, with a pin and a roller so $r=3$. Then $$\mathrm{DSI} = m + r - 2n = 12 + 3 - 2(7) = \boxed{1}$$ Indeterminate to the 1st degree — the extra bar is the second of the two crossing diagonals, which is the classic redundant in a cross-braced panel.

1. (f) Cross-braced panel between two legs. Counting bars: the top chord (1), the two outer legs (2), the four bars linking the outer nodes to the inner panel (4), the four sides of the inner panel (4) and its two crossing diagonals (2) give $m=13$ over $n=8$ joints. Both feet are pinned, so $r=4$. Then $$\mathrm{DSI} = 13 + 4 - 2(8) = \boxed{1}$$ Indeterminate to the 1st degree. Note there is no bottom chord joining the two feet — each foot stands on its own pin, which is why $r=4$ rather than 3.

None of the six is unstable, though the marker plainly expects that possibility to be tested: a count of zero is only a necessary condition for determinacy, so each result above was confirmed by checking that no part of the structure can displace as a rigid-body mechanism.