Question 1 of 8: Determinacy, indeterminacy and stability of six structures

Question 1: Determinacy, indeterminacy 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. (a) A beam on three rollers with a built-in right-hand end and two internal hinges — the figure labels one of them "typical hinge", so both circles on the beam line are hinges. (b) Two beams, an upper one on a pin at its left end and a roller at its right end and a lower one on a pin at its left end and a roller at its right end, joined to each other by a pin link near mid-length. (c) A single-bay frame three storeys high on two fixed feet: the roof beam is rigidly attached to the columns while both floor beams are pin-ended (four circles). (d) A two-bay gabled frame on three fixed feet, the gable rafter continuous with the outer columns and the tie beam pinned to both outer columns. (e) A parallel-chord truss, pin at the left and roller at the right, with X-bracing in the two end panels. (f) A truss occupying a square with the top-right corner cut off, pin at the bottom left and roller at the bottom right; diagonals are not connected where they cross.

Find. For each structure, the classification — unstable, statically determinate, or statically indeterminate — and, when indeterminate, the degree.

Approach. Count restraints against available equations with $i = 3m + r - 3j - c$ for flexural structures and $i = m + r - 2j$ for pin-jointed trusses, then inspect the arrangement, because a favourable count proves only that enough restraints exist, never that they are usefully placed.

1. State the two counting rules. For a plane structure built of flexural members every member carries three internal actions and every joint supplies three equations, so $$i = 3m + r - 3j - c$$ in which $m$ is the number of members, $j$ the number of joints (support points included), $r$ the number of independent reaction components and $c$ the number of released equations of condition — one for each internal hinge that connects two members. For a pin-jointed truss every member carries one unknown force and every joint supplies two equations, so $$i = m + r - 2j$$ A negative $i$ means a mechanism; $i = 0$ means determinate provided the arrangement is sound; $i > 0$ is the degree of indeterminacy.
2. Part (a) — continuous beam, three rollers, one fixed end, two hinges. The rollers supply one component each and the built-in end supplies three, so $r = 3(1) + 3 = 6$. A single straight beam needs three equilibrium equations and each of the two internal hinges releases one moment, so $$i = r - 3 - c = 6 - 3 - 2 = \boxed{1}$$ The arrangement is sound: the built-in end restrains the beam horizontally and rotationally, the piece between the two hinges is carried at both ends, and no two supports coincide. The beam is statically indeterminate to the first degree.
3. Part (b) — two beams joined by a pin. Read the figure as two separate members, one above the other, each carrying a pin at one end and a roller at the other, connected at mid-length by a pin. Reactions total $r = 2 + 1 + 2 + 1 = 6$; the connecting pin transmits two force components; and there are two rigid bodies, hence six equilibrium equations. Therefore $$i = (6 + 2) - 3(2) = \boxed{2}$$ Each beam is already stable on its own pin-and-roller pair, so the link between them is pure surplus and the assembly is indeterminate to the second degree. (Were the small circle read instead as a roller or two-force link transmitting only a vertical force, the count would give $i = 1$; the drawing shows a single hinge circle touching both beams, so two components are adopted.)
4. Part (c) — three-level single-bay frame. Cut the columns at every beam level: the left column becomes three segments, so does the right, and there are three beams, giving $m = 9$ and $j = 8$ (two feet plus three joints on each column). The two fixed feet give $r = 6$. Both floor beams are pinned at each end, which releases four moments, so $c = 4$ and $$i = 3(9) + 6 - 3(8) - 4 = 27 + 6 - 24 - 4 = \boxed{5}$$ The same answer follows from the ring count: three closed rings at three redundants each, less the four moment releases. The frame is indeterminate to the fifth degree.
5. Part (d) — two-bay gabled frame. Segmenting at every joint gives two pieces of each outer column, one middle column, two pieces of tie beam and two rafters, so $m = 9$ and $j = 9$. Three fixed feet give $r = 9$, and the tie beam is pinned to each outer column, so $c = 2$: $$i = 3(9) + 9 - 3(9) - 2 = \boxed{7}$$ The frame is indeterminate to the seventh degree.
6. Part (e) — parallel-chord truss with X-braced end panels. Count the members: four top chords, four bottom chords, five verticals and six diagonals (two in each end panel, one in each central panel), so $m = 19$. There are ten joints and the pin plus roller give $r = 3$: $$i = m + r - 2j = 19 + 3 - 2(10) = \boxed{2}$$ Both surplus members are internal — the second diagonal in each X-braced panel — so the truss is internally indeterminate to the second degree while its reactions remain determinate.
7. Part (f) — count first, then look at the arrangement. The figure has seven joints (three along the bottom, one at each mid-height, two along the top) and eleven members: three edges of the square broken at the mid-height and mid-width joints (six members), the cut-off corner member, the four sides of the central rhombus, less the duplicate already counted — explicitly, top, upper-left, lower-left, bottom-left, bottom-right, right, the cut corner, and the four rhombus sides, plus the long corner-to-corner diagonal. With $r = 3$, $$i = 11 + 3 - 2(7) = 0$$ so the count says determinate.
8. Part (f) continued — test the arrangement and find the mechanism. The four joints of the central rhombus are each free in exactly the direction the rhombus needs: the two mid-height joints lie in the run of the vertical edges, so nothing restrains them horizontally, and the top and bottom mid-width joints lie in the run of horizontal edges, so nothing restrains them vertically. Give the rhombus a virtual mode in which the mid-height joints move $\pm\delta$ horizontally and the top and bottom joints move $\mp\delta$ vertically; every rhombus side then rotates without changing length, because its two end movements are equal and perpendicular to it. The long diagonal joins two corners that do not move at all and, by the note on the figure, is not connected where it crosses the rhombus, so it cannot stop the mode. The structure is therefore $\boxed{\text{unstable}}$ — determinate by count, a mechanism by arrangement.