Question 1 of 8: Stability and determinacy of seven structures

Question 1: Stability and determinacy of seven structures (7 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. Seven plane assemblies. Structures (a) to (e) are built from flexural (beam-type) members, so each member end carries axial force, shear and moment unless a release says otherwise; (f) and (g) are pin-jointed trusses in which the crossing diagonals pass one another without being connected, so each crossing diagonal is a separate two-force member.

Find. For each structure, one of “unstable”, “statically determinate” or “statically indeterminate to degree n”.

[Figure not reproduced: Question 1 — the seven structures, redrawn from the examination paper. (a)–(e) are beam-type assemblies; (f) and (g) are pin-jointed trusses whose crossing diagonals are not connected where they cross. Hatching denotes a fixed base, a triangle on hatching a pin, a triangle on rollers a r. See the official exam paper.]

Approach. Count first, then look: for a beam-type assembly use $i = 3m + r - 3j - c$ and for a pin-jointed truss use $i = m + r - 2j$, then confirm that the members and reactions are actually arranged so as to restrain every degree of freedom, because a satisfactory count proves nothing on its own.

1. State the two counting rules. For a plane assembly of flexural members, $$i = 3m + r - 3j - c$$ where $m$ is the number of members, $j$ the number of joints (support points included), $r$ the number of reaction components and $c$ the number of released conditions (one per moment release in a two-member joint; $n-1$ if a pin joins $n$ members). For a pin-jointed truss the corresponding count is $$i = m + r - 2j .$$ A negative $i$ means a mechanism; $i = 0$ with a sound arrangement means determinate; $i > 0$ means indeterminate to that degree.
2. (a) Fixed end, one internal hinge, two rollers. The reaction components are $r = 3 + 1 + 1 = 5$ and the hinge supplies one condition equation, so $$i = 5 - 3 - 1 = \boxed{1}$$ Horizontal restraint comes from the fixed end and travels through the hinge as an axial force, so the beam is properly restrained: statically indeterminate to the first degree.
3. (b) Beam carried on the apex pin of a two-legged fixed-base frame. Each leg is built in at its base, so each leg on its own already fixes the apex; pinning the two legs together there adds two redundant constraints. The beam, however, touches the rest of the structure only at that one pin, and a pin transmits no moment. Taking moments about the apex for the beam alone leaves its rotation completely unrestrained, so the beam can spin about the apex: the assembly is unstable. (The gross count, ten constraints against nine degrees of freedom, gives $+1$; that is two redundancies in the legs less one mechanism in the beam, and is exactly the case where a count must not be trusted.)
4. (c) Beam on two inclined pin-ended links and two fixed-base columns. Counting the beam in three segments, $m = 7$, $j = 8$, $r = 2 + 2 + 3 + 3 = 10$ and $c = 4$ (one moment release where each link and each column meets the beam), so $$i = 3(7) + 10 - 3(8) - 4 = \boxed{3}$$ The check is quicker the other way round: the beam pinned to two fixed-base columns is already a rigid, once-redundant frame, and each of the two inclined two-force links adds one more constraint. Indeterminate to the third degree.
5. (d) Upper beam bearing on a lower beam. The upper beam has a pin and a roller onto the lower beam, so three interface unknowns; the lower beam has a pin and a roller to ground, so three external unknowns. Six unknowns against six equations (three per beam) gives $$i = 6 - 6 = \boxed{0}$$ The upper beam is solved first and its interface forces then load the lower beam: statically determinate.
6. (e) Beam hung from two inclined links. Ground, two links and the beam form a four-bar linkage. With $m = 3$, $j = 4$, $r = 4$ and $c = 2$, $$i = 3(3) + 4 - 3(4) - 2 = -1$$ so there is one degree of freedom left: the beam can swing. Symmetric loading happens to be in equilibrium, but the structure is a mechanism and is unstable.
7. (f) Trapezoidal truss with crossed diagonals in the centre panel. Counting members: three bottom-chord panels, one top chord, two end inclined members, two verticals and the two unconnected diagonals give $m = 10$; there are $j = 6$ joints and $r = 3$, so $$i = 10 + 3 - 2(6) = \boxed{1}$$ Every panel is triangulated, so the arrangement is sound: indeterminate to the first degree — the second diagonal of the centre panel is the redundant.
8. (g) Two-panel rectangular truss on two pins. Here $m = 8$ (two top-chord panels, two bottom-chord panels, the middle and right verticals, and the two diagonals; there is no member on the left face between the two supports), $j = 6$ and $r = 4$, so $$i = 8 + 4 - 2(6) = \boxed{0}$$ Stability has to be demonstrated because the right-hand panel carries no diagonal of its own. Build the truss up joint by joint from the two pinned joints: the upper middle joint is fixed by the upper-left chord and the diagonal from the lower-left pin; the lower middle joint then follows from the bottom chord and the middle vertical; the lower right joint follows from the bottom chord and the long diagonal from the upper-left pin; and the upper right joint follows from the top chord and the right vertical. All eight members and all four joints are used exactly once, so the truss is rigid: statically determinate.

Two of the seven are therefore mechanisms and, in an examination, they are the marks most often lost: both (b) and (e) pass or nearly pass a numerical count and fail on arrangement. Writing one sentence that names the free motion — “the beam rotates about the apex pin”, “the linkage swings” — is what earns the mark.