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16-Civ-A1 Elementary Structural Analysis · December 2018

Question 1 of 8: Classify each structure — unstable / determinate / indeterminate

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

Notes on this paper

Paper: National Exams — December 2018, 16-Civ-A1 Elementary Structural Analysis (closed book, 3 h; approved Casio/Sharp calculator permitted). Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy (Ch. 2), method of joints & sections (Ch. 3), shear & moment diagrams (Ch. 4), influence lines (Ch. 6), virtual-work deflections (Ch. 8–9), moment distribution (Ch. 11); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged & indeterminate frames.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside), hogging negative; member axial force tension positive (T), compression negative (C).

These A1 papers are defined entirely by their figures, so every structure is redrawn to scale below.

Question 1: Classify each structure — unstable / determinate / indeterminate (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 structures (a)–(d) are beam/frame assemblies; (e)–(f) are pin-jointed trusses (crossing diagonals are not connected). Find. For each, state whether it is unstable, statically determinate, or statically indeterminate, giving the degree of indeterminacy.

Approach. For a beam/frame the degree of static indeterminacy is $\text{DSI}=3m+r-3n-c$ (equivalently $r-3-c$ for a single unbranched load path), where $m$ = members, $n$ = joints, $r$ = reaction components and $c$ = internal condition equations (an internal hinge joining $k$ members releases $k-1$ moments). For a pin-jointed truss $\text{DSI}=m+r-2n$. Positive $\Rightarrow$ indeterminate to that degree; zero $\Rightarrow$ determinate; a negative count (or a geometrically inadequate reaction layout) $\Rightarrow$ unstable.

[Figure not reproduced: (a) beam, 2 internal hinges (b) Π-frame, fixed feet, 1 hinge (c) two-storey frame, 4 hinges (d) Π-frame, pinned feet (e) X-braced truss (f) truss Question 1 — the six structures, redrawn to scale from the exam figure. See the official exam paper.]

CaseCount ($m,\,r,\,n,\,c$)DSIClassification
(a) beampin + 3 rollers ⇒ $r{=}5$; two internal hinges $c{=}2$$r-3-c=5-3-2=0$Statically determinate
(b) Π-frame2 fixed feet $r{=}6$; $m{=}4,\,n{=}5$; one crown hinge $c{=}1$$3(4)+6-3(5)-1=2$Indeterminate, 2°
(c) two-storey2 fixed feet $r{=}6$; $m{=}6,\,n{=}6$; four beam hinges $c{=}4$$3(6)+6-3(6)-4=2$Indeterminate, 2°
(d) Π-frame2 pinned feet $r{=}4$; $m{=}3,\,n{=}4$; no hinge $c{=}0$$3(3)+4-3(4)-0=1$Indeterminate, 1°
(e) X-truss$m{=}12,\,n{=}7$; pin + roller $r{=}3$$m+r-2n=12+3-14=1$Indeterminate, 1°
(f) truss$m{=}7,\,n{=}6$; two pins $r{=}4$$m+r-2n=7+4-12=-1$Unstable (a mechanism)
  1. (a) two-hinge beam. A pin plus three rollers give $r=5$ reaction components; the two internal hinges supply $c=2$ extra equations, so $\text{DSI}=5-3-2=\boxed{0}$ — a determinate compound beam.
  2. (b) vs (d) — the same Π-frame, different feet. With fixed feet ($r=6$) and a single crown hinge, $\text{DSI}=3(4)+6-3(5)-1=2$. With pinned feet ($r=4$) and no hinge the same trapezoidal portal gives $3(3)+4-3(4)=1$. A built-in-both-ends member is 3° redundant; each pin foot and each internal hinge peels off one redundancy.
  3. (c) two-storey frame. Fixed feet ($r=6$); each of the two floor beams carries two hinges ($c=4$ total). Modelling the four beam-and-column segments meeting at the rigid corners, $3(6)+6-3(6)-4=2$.
  4. (e) X-braced panel. The central panel carries both crossing diagonals, so it is over-braced by one bar: $m+r-2n=12+3-14=1$. The joint-equilibrium matrix has full rank ($=2n$), confirming the surplus is a genuine redundancy, not a mechanism.
  5. (f) truss. Seven bars and two pins give $m+r=11$ against $2n=12$ equilibrium equations — one equation short. The reaction/member layout cannot equilibrate a general load: the structure is unstable (a mechanism). The joint matrix is rank-deficient, confirming it.
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