Question 1 of 8: Stability and Degree of Indeterminacy
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 07-Str-A1 Elementary Structural Analysis, National Examinations, May 2018. Three hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: Questions 1 to 5 are compulsory and one of Questions 6, 7 or 8 is chosen. Marks are shown against each question and total 100. Every question is worked here, including all three of the optional Questions 6, 7 and 8, so that the set can be used as a study resource.
Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 stability and determinacy; Ch. 4 influence lines; Ch. 6 deflections by virtual work; Ch. 11 displacement method; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 8, 13, 16); K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods. For Canadian design practice the companion documents are the National Building Code of Canada (Part 4, load combinations) and CSA S16 Design of Steel Structures; this paper is pure analysis, so no design code is invoked in the answers.
Check — readings taken from the printed figures. Every dimension, support symbol and load extent below was read from the printed figures. Four readings carry the answers and are stated explicitly so that a marker can check them: (i) in Question 2(a) the 10 kN/m block covers only the 2 m overhang, ending at the roller; (ii) in Question 4(a) the truss is a Warren truss whose top nodes are offset 3 m from the bottom nodes (the 3 m dimension at the top right confirms this); (iii) in Question 7 the 10.8 kN/m ruling starts directly above joint 1 and runs to joint 4, so it acts over the whole horizontal projection, inclined member included; (iv) in Question 8 the thin vertical line between U2 and L3 touches neither joint circle and is the witness line of the 3 m dimension, not a member — taking it as a member would make the truss statically indeterminate and the question unanswerable by virtual work.
Question 1: Stability and Degree of Indeterminacy (6 marks)
Given. Six planar structures: (a) a beam on a pin and two rollers with one internal hinge; (b) a loaded beam pinned at each end to the head of an inclined leg, the two legs meeting at a pinned apex, with a pin and a roller under the beam itself; (c) a trapezoidal frame whose horizontal member is pin-connected to two inclined legs with fixed bases; (d) a Z-shaped frame with rigid corners on a pin, a pin and a roller; (e) a two-panel truss with crossing (unconnected) diagonals in both panels, pinned to a wall at both left-hand joints; (f) a Warren truss pinned to a wall at its upper left joint and propped by a single inclined bar from a second wall pin.
Find. A one-word classification for each structure — unstable, statically determinate, or statically indeterminate to degree i.
[Figure not reproduced: Question 1: the six structures, redrawn from the examination paper. Open circles are internal hinges (or, in the trusses, pinned joints); hatching denotes a fixed base, a triangle a pin and a triangle on rollers a roller. See the official exam paper.]
Approach. Count the unknowns against the available equations — for a rigid-jointed structure $i = (3m + r) - (3j + c)$ and for a pin-jointed truss $i = (m + r) - 2j$ — and then, because a count alone cannot see a mechanism, partition each structure and ask what motion is still free.
Fix the counting rules and the trap they hide. For beam and frame members the general count is
$$i \;=\; (3m + r) \;-\; (3j + c)$$
in which $m$ is the number of members, $r$ the number of reaction components, $j$ the number of joints (support joints included) and $c$ the number of released conditions (one per hinge connecting two members). For a pin-jointed truss the equivalent statement is $i = (m + r) - 2j$. A positive $i$ means indeterminate, zero means determinate provided the arrangement is proper, and a negative value means a mechanism. The count is necessary but never sufficient: three reactions whose lines of action are concurrent or parallel give $i = 0$ and still collapse.
Structure (a) — pin, two rollers, one internal hinge. The reactions are two components at the pin and one at each roller, so $r = 4$, and the single hinge supplies one condition equation in addition to the three equations of global equilibrium. Hence $$i = r - (3 + c) = 4 - (3 + 1) = 0.$$ The arrangement is proper: the portion to the right of the hinge is carried by two vertical rollers and by the hinge force, and the pin at the far left supplies the horizontal restraint that the rollers cannot. The structure is $\boxed{\text{statically determinate}}$.
Structure (b) — beam on two inclined legs. Each leg is pinned at both ends and carries no load along its length, so each is a two-force member and contributes one unknown, its axial force. The unknowns are therefore the two leg forces, the two components of the apex pin reaction and the three components supplied by the pin and roller under the beam — seven in all. The equations available are the three for the beam as a free body plus the two force equations at the apex node, five in total, so
$$i = 7 - 5 = 2.$$
The structure is $\boxed{\text{indeterminate to the 2nd degree}}$. Note that the two supports sit on the beam itself, not at the truss nodes; reading them at the nodes changes the count.
Structure (c) — trapezoidal frame, fixed bases, pinned knees. Here $m = 3$, $j = 4$, $r = 3 + 3 = 6$ and the two knee pins each release one moment, so $c = 2$:
$$i = (3 \times 3 + 6) - (3 \times 4 + 2) = 15 - 14 = 1,$$
that is, $\boxed{\text{indeterminate to the 1st degree}}$. The same result follows from the two-hinged-arch view: six support unknowns against three equilibrium equations and two hinge conditions.
Structure (d) — Z-frame with rigid corners. There are no releases, so only the reactions matter: two at the upper-left pin, two at the pin below the vertical member and one at the right-hand roller, giving $r = 5$ and
$$i = r - 3 = 5 - 3 = 2,$$
so the frame is $\boxed{\text{indeterminate to the 2nd degree}}$.
Structure (e) — two panels, both cross-braced. Counting the members: two top chords, two bottom chords, one right-hand vertical and four diagonals give $m = 9$; there are $j = 6$ joints and two wall pins give $r = 4$. Then
$$i = (m + r) - 2j = (9 + 4) - 12 = 1.$$
The redundancy is real and internal — the right-hand panel is a quadrilateral carrying both of its diagonals — so the truss is $\boxed{\text{indeterminate to the 1st degree}}$. There is no vertical member between the two wall joints and none at mid-length; inventing either changes the answer.
Structure (f) — Warren truss on a wall pin and an inclined bar. The inclined bar from the lower wall pin is loaded only at its ends, so it acts as a link and contributes one reaction component along its own axis. Excluding it from the member list gives $m = 9$ (two top chords, two bottom chords and five web diagonals), $j = 6$ and $r = 2 + 1 = 3$, hence
$$i = (9 + 3) - 2(6) = 0.$$
Stability follows constructively: the upper wall joint is a fixed point, the first bottom joint is held by that joint through one diagonal and by the link along a different line, and every remaining joint is then fixed by two more non-collinear bars. The truss is $\boxed{\text{statically determinate}}$.
Two habits are worth carrying out of this question. First, after computing $i$, partition the structure and ask of each part what motion is still free — that is the only way to catch a sub-assembly that is over-restrained while another is a mechanism, a pairing that cancels in the total. Second, never let the reaction count alone classify a truss: the internal arrangement decides as often as the supports do, as structures (e) and (f) show.
