Question 1 of 8: Determinacy and stability of six structures

Question 1: Determinacy and stability of six structures (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, each carrying a nominal load that plays no part in the classification. Read from the drawings: (a) a continuous beam on three rollers with a built-in right-hand end and two internal hinges; (b) a portal frame on two pinned bases with a hinge at the right knee; (c) a two-storey, two-bay frame on three fixed bases with four hinges in the roof beam; (d) two beams tied by two posts, all joints rigid, carried on one pin and three rollers; (e) a two-panel truss on a pin and a roller whose central diagonals cross without connecting; (f) a two-storey braced frame on three pinned bases.

Find. For each structure, whether it is unstable, statically determinate or statically indeterminate, and in the last case the degree of static indeterminacy.

[Figure not reproduced: The six structures as drawn on the exam paper. Open red circles are internal hinges; a plain triangle on hatching is a pin, a triangle riding on rollers is a roller, and bare hatching with the member running into it is a built-in (fixed) end. See the official exam paper.]

Approach. Count the unknowns and the available equations — for a beam or frame the degree is $i = 3m + r - 3j - c$, and for a pin-jointed truss it is $i = m + r - 2j$ — and then, because a positive count proves nothing on its own, test each structure part by part for a motion that the restraints leave free.

1. (a) Continuous beam: three rollers, a built-in end and two hinges. The three rollers supply one vertical component each and the built-in end supplies three, so $r = 3(1) + 3 = 6$. Statics gives three equations and each internal hinge adds one condition equation ($\sum M = 0$ about the hinge for the part beyond it), so $$i = r - (3 + c) = 6 - (3 + 2) = \boxed{1}$$ Stability follows by walking the beam: the left portion is carried on two rollers and pinned to the middle link, the middle link between the hinges is supported at both ends, and the horizontal restraint the rollers cannot supply is provided by the built-in end through the axial force in the beam. The structure is statically indeterminate to the first degree.
2. (b) Portal frame: two pinned bases and one knee hinge. Each pin supplies two components, so $r = 4$, and the single hinge at the beam-to-column joint on the right adds one condition: $i = 4 - (3 + 1) = 0$. The arrangement is the classical three-hinged frame, and it is stable because the three hinges — the two bases and the knee — are not collinear. Note that the left knee is drawn rigid and the right knee carries a small open circle; reading them the other way round changes nothing here, but reading the bases as one pin and one fixed end would make the count $6 - 4 = 2$ and the answer wrong. The frame is statically determinate.
3. (c) Two-storey, two-bay frame on three fixed bases. Take the three columns, each broken into a lower and an upper length at the intermediate beam, plus the two halves of each beam: $m = 10$, $j = 9$, $r = 3(3) = 9$, and the roof beam carries four hinges, so $$i = 3m + r - 3j - c = 3(10) + 9 - 3(9) - 4 = \boxed{8}$$ The same number follows from the closed-ring rule: the frame encloses four panels, each worth three redundants, less the four releases, $3(4) - 4 = 8$. It is statically indeterminate to the eighth degree.
4. (d) Twin beams tied by two posts. The two posts stand between the beams at interior points only — the ends of the beams are not connected to each other — so the assembly encloses exactly one closed panel. With $m = 8$, $j = 8$ and $r = 2 + 3(1) = 5$, $$i = 3(8) + 5 - 3(8) - 0 = \boxed{5}$$ which is the closed panel ($3$) plus the two surplus external components ($5 - 3$). The single pin restrains the assembly horizontally and the four vertical components are neither all parallel through one point nor concurrent, so it is statically indeterminate to the fifth degree.
5. (e) Two-panel truss with crossing diagonals. Counting the members off an enlargement gives $m = 14$ on $j = 8$ joints, with a pin and a roller, $r = 3$: $$i = m + r - 2j = 14 + 3 - 16 = \boxed{1}$$ Both central diagonals are present and the note "diagonals are not connected where they cross" means they act as two independent bars, which is precisely the one redundancy. Forming the $16 \times 17$ joint-equilibrium matrix and checking its rank confirms that all sixteen equations are independent, so the truss is stable: statically indeterminate to the first degree.
6. (f) Two-storey braced frame on three pins. Here $m = 11$, $j = 8$ and $r = 3(2) = 6$, giving $i = 11 + 6 - 16 = \boxed{1}$. This case rewards a second look, because the external count alone ($r - 3 = 3$) suggests a much higher degree; the frame is externally over-restrained by three and internally short of members by two, and the two effects very nearly cancel. The rank check again returns the full sixteen, so nothing is left free to move and the frame is statically indeterminate to the first degree.