07-Str-A1 · December 2018

Question 1 of 8: Stability and Determinacy of Six Structures

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2018 — 07-Str-A1 Elementary Structural Analysis; 3 hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: answer all of Questions 1–5, then one only of Questions 6, 7 or 8. Marks are shown in the left margin (6, 18, 18, 18, 18, then 22 for the optional question). Every question is worked below, including all three alternatives, because the full set is the more useful study resource.

Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 determinacy and stability; Ch. 4 influence lines; Ch. 6–9 deflections and virtual work; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 6, 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 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 — two figure readings that the printed figures leave open, both stated where they are used. (i) In Question 2(c) the hatched plane under the upper roller is drawn parallel to the inclined member, so the reaction is taken normal to the member. That reading makes every result an integer (R = 28 kN, shears 32 / −4 / −28 kN, axial forces 25 / 10 / 0 kN), whereas a vertical-plane reading gives nothing round; and because the two readings differ only by a force acting along the member axis, the shear and moment diagrams are identical either way. (ii) In Question 7 both wall supports are drawn as triangles on hatching. The truss is analysable only if one of them releases a component; taking the pin at U₂ and the horizontal-only roller at L₃ gives δ = 29.0 mm, and the same value follows if the thin vertical line between the two wall joints is a dimension witness line rather than a member. The alternative (pin at L₃) would give 31.8 mm.

Question 1: Stability and Determinacy 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 planar structures: (a) a continuous beam on a pin and three rollers carrying a full-length UDL, with two internal hinges (labelled TYPICAL HINGE) in the bay between the second and third supports; (b) a trapezoidal frame with rigid knees, fixed bases and one internal hinge at mid-beam; (c) a two-storey single-bay frame, columns fixed at their bases, rigid beam-to-column joints, and two internal hinges in each of the two loaded beams; (d) the same trapezoid as (b) but with pinned bases and no hinge; (e) a triangulated truss with a pin and a roller, whose central panel carries both of its diagonals (not connected where they cross); (f) an irregular truss on two pinned bases carrying two horizontal loads.

Find. The classification of each structure, with the degree of indeterminacy where it applies.

(a) continuous beam — pin + 3 rollers, 2 internal hinges

(b) fixed bases, rigid knees, 1 hinge at mid-beam

Question 1, structures (a) and (b).

(d) pinned bases, rigid knees, no hinge

Question 1, structures (c) and (d).

(e) truss — pin + roller, central panel double-diagonalled

(f) truss — two pinned bases, horizontal loads

Question 1, structures (e) and (f).

Approach. Count first, then interrogate the arrangement: for beam-type and framed structures use $i = 3m + r - 3j - c$ and for pin-jointed trusses $i = m + r - 2j$, and afterwards ask of every partitioned piece "what motion is still free?", because a favourable count never by itself proves stability.

Part (a) — a Gerber beam: count the release equations, then check that the suspended span is genuinely carried. One continuous member, $r = 2 + 1 + 1 + 1 = 5$ and $c = 2$, so $$i = r - 3 - c = 5 - 3 - 2 = \boxed{0}$$ The count alone is not enough here, because the two hinges bracket a length of beam that carries no support of its own. Writing the four independent equations (global $\Sigma F_y$ and $\Sigma M$, plus zero moment at each hinge) and solving symbolically gives a non-singular coefficient matrix, so the piece between the hinges is a suspended span carried by the two cantilevered ends — the classic Gerber arrangement. Statically determinate.
Part (b) — an open frame with fixed bases and one release. $m = 3$ (leg, beam, leg), $j = 4$, $r = 3 + 3 = 6$, $c = 1$, hence $$i = 3(3) + 6 - 3(4) - 1 = 9 + 6 - 12 - 1 = \boxed{2}$$ Statically indeterminate to the second degree. The hinge buys one equation back from the three redundants a two-fixed-base open frame would otherwise carry.
Part (c) — two closed panels, four releases. Taking joints at both bases, both mid-height beam-to-column intersections and both roof corners gives $j = 6$ and $m = 6$ (two column lengths each side, two beams); $r = 6$ and $c = 4$: $$i = 3(6) + 6 - 3(6) - 4 = 18 + 6 - 18 - 4 = \boxed{2}$$ Each closed rectangle contributes three redundants and each hinge removes one: $3 + 3 = 6$ against four hinges. Statically indeterminate to the second degree.
Part (d) — the same geometry with pins instead of fixed bases and no hinge. $m = 3$, $j = 4$, $r = 2 + 2 = 4$, $c = 0$: $$i = 3(3) + 4 - 3(4) - 0 = \boxed{1}$$ Statically indeterminate to the first degree — the familiar two-hinged arch or portal. Comparing (b) with (d) is the point of the pair: releasing both base moments removes two redundants, adding the crown hinge removes one more, and only the order in which you remove them changes.
Part (e) — a truss whose central panel carries two diagonals. Counting the members gives six bottom-chord bars, one horizontal top chord, the two rafter segments each side of the apex, two verticals and two crossing diagonals, i.e. $m = 12$ over $j = 7$ joints with $r = 3$: $$i = m + r - 2j = 12 + 3 - 14 = \boxed{1}$$ The redundancy is easy to name: the quadrilateral panel bounded by the two verticals, the horizontal top chord and the bottom chord needs only one diagonal to be rigid, and it is given both. Statically indeterminate to the first degree.
Part (f) — the deliberately deficient one. The bars are $A\!-\!C$, $C\!-\!E$, $E\!-\!D$, $D\!-\!B$, $C\!-\!D$, $E\!-\!F$ and $F\!-\!B$, so $m = 7$, $j = 6$ and $r = 2 + 2 = 4$: $$i = m + r - 2j = 7 + 4 - 12 = \boxed{-1}$$ A negative index means the assembly is a mechanism, and the motion is easy to identify: the four bars $A\!-\!C$, $C\!-\!D$, $D\!-\!B$ with pins at $A$ and $B$ form a four-bar linkage with one degree of freedom; joint $E$ is then tied rigidly to bar $C\!-\!D$ by the triangle $C\!-\!D\!-\!E$, and joint $F$ is located by its two bars to $E$ and $B$. Counting freedoms instead of bars says the same thing — four movable joints give eight degrees of freedom against only seven bar constraints. Unstable.

Structure	m	r	j	c	Index	Classification
(a) continuous beam	1 member	5	—	2	0	Statically determinate
(b) trapezoidal frame, fixed bases	3	6	4	1	+2	Indeterminate, 2nd degree
(c) two-storey frame	6	6	6	4	+2	Indeterminate, 2nd degree
(d) trapezoidal frame, pinned bases	3	4	4	0	+1	Indeterminate, 1st degree
(e) truss with crossed diagonals	12	3	7	—	+1	Indeterminate, 1st degree
(f) irregular truss, two pins	7	4	6	—	−1	Unstable (1 DOF mechanism)

Topic. Static determinacy and stability of planar structures

Key relations. $i = 3m + r - 3j - c$ for framed structures and $i = m + r - 2j$ for pin-jointed trusses; $i < 0$ means a mechanism, $i = 0$ determinate, $i > 0$ indeterminate to degree $i$. Condition equations: an internal hinge connecting $n$ members releases $n - 1$ moments.

Why this works. The index compares the number of unknown force quantities with the number of independent equilibrium and condition equations available. It is a necessary but not a sufficient test, because it counts equations without asking whether they are independent. Three reaction lines that are parallel, or that meet at a common point, give a singular coefficient matrix even when the count is zero, and the same happens internally when a panel is left without a diagonal while another panel is given two. The reliable habit is therefore two-part: compute the index, then partition the structure and ask what rigid-body motion each part still has.

Common pitfall. Classifying a truss from $r - 3$ alone. Structure (f) has four reaction components and would read as externally indeterminate on that basis, yet it is a mechanism because it is one bar short internally; the member count and the reaction count have to be taken together.

Source. Hibbeler, Structural Analysis, §2.4–2.5 (beams and frames) and §3.3 (trusses); Kassimali, Structural Analysis, §3.4 and §4.4.

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