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. (a) A beam on a roller at its left end, an intermediate vertical prop, two internal hinges near the right, and a fixed (built-in) right end, carrying a uniform load w. (b) A rigid L-frame: a pinned column carrying a horizontal beam that also rests on a pin support part way along, with w over the beam. (c) An upper beam built in at the left and resting on a roller, joined by a vertical two-force link to a lower beam that carries only a roller; both beams carry w. (d) A three-storey, single-bay rigid frame on two pinned bases carrying three horizontal loads P. (e) A parallel-chord truss, pinned at one end and on a roller at the other, with one X-braced panel (the crossing diagonals are not connected). (f) A tapered tower truss on two pinned bases with two X-braced panels; the broken lines are the leg axes produced.

Find. For each structure, the classification — unstable, statically determinate, or statically indeterminate to a stated degree.

[Figure not reproduced: Figure 1 — the six structures of Question 1, redrawn from the examination paper. See the official exam paper.]

Approach. Count the unknown reaction components $r$, the equations of condition $c$ contributed by internal hinges and links, and (for the trusses) the members $m$ and joints $j$; then test the arrangement, because a favourable count never proves stability.

For a rigid-body assembly of beam-type members the degree of indeterminacy is

where $c$ is the number of equations of condition (one per internal hinge in a beam, one per two-force link), and for a plane truss it is

A zero or positive count means only that there are enough restraints; it does not say that they are arranged so as to prevent motion. Structure (c) below is exactly that trap, and it is the reason the answer is written out member-by-member rather than as a bare arithmetic total.

1. (a) Beam with a prop and two internal hinges. The reactions are one vertical component at the left roller, one at the intermediate prop and three at the built-in end, so $r = 1 + 1 + 3 = 5$; the two internal hinges give $c = 2$. Hence $$ i = 5 - (3 + 2) = \boxed{0} $$ and the arrangement is sound: the built-in end anchors the chain horizontally through both hinges, the suspended length between the hinges is carried by the two adjacent lengths, and the roller-plus-prop length is held against rotation by the hinge force. The beam is statically determinate.
2. (b) Rigid L-frame on two pins. The column base is a pin (two components) and the beam rests on a second pin (two components), so $r = 4$. There is no internal hinge, so the frame is one rigid body with three equations of equilibrium: $$ i = 4 - 3 = \boxed{1} $$ The two pins are not on a common line through which all loads pass, so the frame is stable. It is statically indeterminate to the first degree.
3. (c) Two beams joined by a vertical link — the planted trap. The count is favourable: three rigid bodies (upper beam, link, lower beam) give nine equations; the unknowns are five reaction components (three at the built-in end, one at each roller) plus two pin forces at each end of the link, which is also nine, so the arithmetic returns $$ i = 5 + 2(2) - 3(3) = \boxed{0} . $$ Now look at the arrangement. The link is pinned at both ends and carries no transverse load, so it is a two-force member and can only push or pull along its own vertical axis. The lower beam therefore receives a vertical force from the link and a vertical force from its roller, and nothing else: its horizontal equilibrium equation degenerates to $0 = 0$. The lower beam is free to translate horizontally, so the assembly is a mechanism with one degree of freedom while being over-restrained by one elsewhere; the two cancel in the total. The structure is unstable.
4. (d) Three-storey single-bay frame on two pins. Take the members as the three beams and the six column lengths between floors, so $m = 9$, with joints at the two bases and at the six beam-column intersections, $j = 8$, and $r = 2 + 2 = 4$. For a rigid plane frame, $$ i = 3m + r - 3j = 27 + 4 - 24 = \boxed{7} $$ which is the familiar result that each of the three closed cells is three times redundant, less the two moment releases introduced by the pinned bases. It is statically indeterminate to the seventh degree.
5. (e) Parallel-chord truss with one X-braced panel. Counting off the drawing: five bottom-chord members, three top-chord members, two end diagonals, four verticals and four inclined web members (two of which cross in the middle panel without being connected) give $m = 18$; there are six bottom and four top joints, $j = 10$; the pin and roller give $r = 3$. Hence $$ i = (18 + 3) - 2(10) = \boxed{1} . $$ Externally $r - 3 = 0$, so all of the redundancy is internal — the second diagonal of the X-braced panel. The truss is statically indeterminate to the first degree.
6. (f) Tapered tower on two pins. The members are the top chord, the intermediate horizontal, four leg lengths and the four diagonals of the two X-braced panels, $m = 10$; the joints are the two at the top, two at mid-height and two at the base, $j = 6$; two pins give $r = 4$. Hence $$ i = (10 + 4) - 2(6) = \boxed{2} , $$ made up of one external redundant ($r - 3 = 1$) and one internal redundant ($m - (2j - 3) = 10 - 9 = 1$, the extra diagonal in the upper panel). The lower panel is a simple two-joint extension and adds nothing. It is statically indeterminate to the second degree.

The broken lines on structure (f) are the leg axes produced to their point of intersection above the truss. They are a geometric construction, not members; on a truss carried by two pins the concurrency of those axes has no bearing on stability. The same drawing on a truss supported by three two-force links would be the classic warning that the reaction lines are concurrent and the assembly can rotate about the meeting point — worth recognising, and worth dismissing here for the right reason.