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. Structures (a) to (d) are built of flexural (beam-type) members that transmit axial force, shear and moment across every joint; structures (e) and (f) are pin-jointed trusses whose members carry axial force only, and their crossing diagonals are not connected where they cross, so each crossing pair is two separate members and no joint exists at the crossing point. The support symbols carry the usual meanings: hatched line = fixed (3 reaction components), triangle on hatching = pin (2), triangle on rollers = roller (1); a small open circle in the run of a member is an internal hinge, which releases the moment there.

Find. For each structure, classify it as unstable, statically determinate or statically indeterminate, and where it is indeterminate give the degree.

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

1. Set out the two counting rules and what they mean. For a plane structure built of flexural members, each member carries three internal actions and each joint supplies three equations, so the degree of static indeterminacy is $$i = 3m + r - 3j - c$$ where $m$ 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 connecting 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 the structure is a mechanism, $i = 0$ means it is statically determinate if it is also properly arranged, and $i > 0$ is the degree of indeterminacy.
2. Part (a) — continuous beam with one internal hinge. The pin contributes two reaction components and each of the two rollers one, so $r = 4$. The three equations of overall equilibrium are supplemented by one condition equation, $\sum M = 0$ taken on either side of the hinge, giving four independent equations for four unknowns: $$i = r - (3 + c) = 4 - (3 + 1) = \boxed{0}$$ Checking the arrangement, the length of beam to the right of the hinge is held by the hinge (two force components) and its own roller (one) — three restraints on one rigid body, none of them parallel or concurrent — and the remainder is then held by the pin and the middle roller. The beam is statically determinate.
3. Part (b) — two-bay frame, fixed / pinned / fixed. Take the members as the two beam spans and the three columns, $m = 5$, and the joints as the three bases and the three beam-to-column intersections, $j = 6$. The two fixed bases give three components each and the pinned base two, so $r = 3 + 2 + 3 = 8$, and there is no internal release, $c = 0$. Hence $$i = 3(5) + 8 - 3(6) - 0 = 15 + 8 - 18 = \boxed{5}$$ The same answer follows from the loop count: the frame encloses two closed circuits with the foundation, worth $3 \times 2 = 6$ redundants, from which the single moment release at the pinned base is subtracted. It is indeterminate to the fifth degree.
4. Part (c) — trapezoidal frame with a crown hinge. The four members are the two inclined legs and the two halves of the top chord, $m = 4$, meeting at five joints (two bases, two knees and the hinge), $j = 5$. Both bases are fixed, $r = 6$, and the crown hinge joins two members, releasing one equation, $c = 1$: $$i = 3(4) + 6 - 3(5) - 1 = 12 + 6 - 15 - 1 = \boxed{2}$$ Counting externally gives the same figure and is quicker: six reaction components against three equations of equilibrium plus one condition equation at the hinge leaves $6 - 4 = 2$. The frame is indeterminate to the second degree. Had the bases been pinned instead of fixed it would have been the familiar determinate three-hinged frame.
5. Part (d) — stepped two-bay frame. The members are the tall left column, the upper beam, the two lengths of the centre column above and below the lower beam, the lower beam, and the short right column, so $m = 6$; the joints are the three bases, the two upper corners and the two lower-beam corners, but the centre column and the lower beam share one joint, so $j = 7$. With a fixed base at each outside column and a pin at the centre, $r = 3 + 2 + 3 = 8$ and $c = 0$: $$i = 3(6) + 8 - 3(7) = 18 + 8 - 21 = \boxed{5}$$ It is indeterminate to the fifth degree — the same as (b), which is worth noticing: stepping the frame changes the geometry but not the number of independent circuits or releases.
6. Part (e) — truss with crossing diagonals. Because the diagonals are not connected where they cross, there is no joint at the crossing and each diagonal is a single member running from chord to chord. The count is two top-chord members, two bottom-chord members, and seven web members (two from the left support, three from the loaded apex and two from the right support), $m = 11$, with $j = 6$ joints. A pin and a roller give $r = 3$: $$i = m + r - 2j = 11 + 3 - 12 = \boxed{2}$$ The two surplus members are the extra diagonals in the two panels; removing one from each panel would leave a determinate, still stable truss. It is indeterminate to the second degree.
7. Part (f) — truss on three links: the count is misleading. The upper assembly is a properly triangulated chain of four triangles, nine members on six joints, and it is carried on three two-force links which run down to three pinned bases. Counting everything, $m = 9 + 3 = 12$, $j = 6 + 3 = 9$ and $r = 6$, so $$m + r = 12 + 6 = 18 = 2j$$ which announces a statically determinate truss. The arrangement contradicts the count. A rigid body held by three links can only be stable if the three link axes are neither all parallel nor all concurrent, and here the broken lines in the figure show that the axis of the inclined left link, the axis of the inclined centre link and the vertical right link all pass through one point. The assembly can rotate through an infinitesimal angle about that point without straining any link, so no set of link forces can equilibrate a moment about it. The structure is $\boxed{\text{unstable}}$ — geometrically (or instantaneously) unstable, and the applied horizontal loads make the deficiency a real one rather than an academic curiosity.
8. Collect the six answers and state the general lesson. Five of the six can be settled by counting alone; the sixth cannot, and it is the one the examiner uses to separate candidates who count from candidates who look. Always finish a determinacy question by asking whether the restraints are arranged so that they can actually resist a general force system.