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 under a uniform load carrying a pin, one internal hinge and two rollers; (b) a single-bay two-storey rigid frame with a uniform load on each beam and both bases pinned; (c) a gable frame of two inclined members meeting at a rigid apex under a point load, one base fixed and the other pinned; (d) a uniformly loaded beam on two end rollers, propped at two interior points by inclined members that run down to ground pins; (e) a two-panel braced frame whose crossing diagonals are connected at the crossing, both bases pinned; (f) a parallel-chord truss with two X-braced interior panels whose diagonals are not connected at the crossings, carrying a pin and a roller.

Find. For each structure, a verdict of unstable, statically determinate or statically indeterminate, with the degree stated in the last case.

[Figure not reproduced: Question 1 — the six structures as printed. Support symbols are read from the drawing: a plain hatched triangle is a pin, a triangle on two circles is a roller, and bare hatching with the member running into it is a fixed end. See the official exam paper.]

Approach. Count unknowns against equations with the two standard forms — $i = 3m + r - 3j - c$ for structures built of flexural (beam-type) members and $i = m + r - 2j$ for pin-jointed (truss-type) members — then, because a count alone can never prove stability, partition each structure and ask what motion is left free.

1. Fix the notation before counting anything. For a rigid-jointed plane structure every member carries three internal actions and every joint supplies three equations, so $$i \;=\; 3m + r - 3j - c$$ where $m$ is the number of members between joints, $j$ counts every joint including the supports, $r$ is the number of reaction components and $c$ is the number of released conditions (one per internal hinge in a two-member joint). For a pin-jointed assembly each bar carries one force and each joint supplies two equations, so $$i \;=\; m + r - 2j.$$ A positive $i$ is the degree of static indeterminacy, $i = 0$ means determinate provided the arrangement is stable, and $i < 0$ is a mechanism.
2. (a) Beam with a pin, one hinge and two rollers. A straight beam needs no member/joint bookkeeping: the restraint count is $r = 2 + 1 + 1 = 4$, the equations available are the three of plane statics plus one condition equation for the single hinge, so $$i \;=\; r - (3 + c) \;=\; 4 - (3 + 1) \;=\; \boxed{0}$$ and the beam is statically determinate. It is also stable: the pin fixes the left end horizontally and vertically, each roller adds one vertical restraint, and the hinge is bracketed by supports on both sides, so no segment is free to rotate.
3. (b) Two-storey single-bay rigid frame, both bases pinned. Number the joints: two bases, two mid-height joints and two roof joints give $j = 6$; the members are four column segments and two beams, $m = 6$; the two pins give $r = 4$. Hence $$i \;=\; 3(6) + 4 - 3(6) - 0 \;=\; \boxed{4}$$ — indeterminate to the fourth degree. The result is easy to check against the closed-loop form: each of the two closed rectangles is three times redundant, giving six, and each pinned base releases one of them, leaving four.
4. (c) Gable frame, rigid apex, one fixed base and one pin. Here $m = 2$, $j = 3$ (two bases and the apex) and $r = 3 + 2 = 5$, so $$i \;=\; 3(2) + 5 - 3(3) \;=\; \boxed{2}$$ — indeterminate to the second degree. This is the sub-part that punishes a careless reading of the supports: the left base has no triangle, only hatching with the member running into it, which is a fixed end. Reading it as a pin would give $r = 4$ and a first-degree answer, and reading the apex as hinged would remove another degree.
5. (d) Beam on two end rollers propped by two inclined members. The beam is divided by the two prop connections into three segments, and each prop is a member, so $m = 5$; the joints are the two beam ends, the two prop connections and the two ground pins, $j = 6$; the restraints are two rollers and two pins, $r = 1 + 1 + 2 + 2 = 6$. Then $$i \;=\; 3(5) + 6 - 3(6) \;=\; \boxed{3}$$ — indeterminate to the third degree. The count is the closed circuit formed by the beam, the two props and the ground; that ring alone accounts for all three.
6. (e) Two-panel braced frame, diagonals connected at the crossings. Because the diagonals are connected where they cross, each crossing is a real joint and each diagonal is two bars. Counting: six frame joints plus two crossing joints give $j = 8$; the members are four column segments, two horizontal chords and eight diagonal halves, $m = 14$; the two pinned bases give $r = 4$. Then $$i \;=\; m + r - 2j \;=\; 14 + 4 - 16 \;=\; \boxed{2}$$ — indeterminate to the second degree. Each fully cross-braced panel supplies one redundant diagonal, and there are two panels.
7. (f) Parallel-chord truss, diagonals NOT connected at the crossings. The instruction that the crossing diagonals are not connected is what keeps the crossings out of the joint count. With five bottom joints and five top joints, $j = 10$; the members are four bottom chords, four top chords, three verticals, two end diagonals, two second diagonals and the four bars of the two X-panels, $m = 19$; the pin and roller give $r = 3$. Hence $$i \;=\; 19 + 3 - 20 \;=\; \boxed{2}$$ — indeterminate to the second degree, one redundant diagonal per X-panel, exactly as in (e).
8. Confirm stability, because a count never does. Structure (b) is a ring, (c) has a fixed base and a pin, (d) is a triangulated circuit through the ground, and (e) and (f) are fully triangulated with adequate external restraint, so all of them are stable and their positive counts are genuine degrees of redundancy. Structure (a) is the only determinate one; its hinge is flanked by supports, so no rigid body within it can move. None of the six is a mechanism.