16-Civ-A1 Elementary Structural Analysis · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper: National Exams — May 2018, 16-Civ-A1 Elementary Structural Analysis (closed book, 3 h; 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 / slope–deflection (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, 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 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.
Approach. For beam/frame assemblies the degree of static indeterminacy is $\text{DSI}=3m+r-3n-c$ (equivalently $r-3-c$ for a tree-like load path with no closed ring), where $m$ = members, $n$ = joints, $r$ = reaction components and $c$ = internal condition equations (one released moment per internal hinge). For pin-jointed trusses $\text{DSI}=m+r-2n$. Positive $\Rightarrow$ indeterminate to that degree; zero $\Rightarrow$ determinate; negative (or a geometrically inadequate reaction layout) $\Rightarrow$ unstable.
[Figure not reproduced: (a) (b) (c) (d) (e) (f) Q1 — the six structures, redrawn from the source (circles mark internal hinges / pin joints). See the official exam paper.]
| Case | Reactions $r$ / members $m$ / joints $n$ / releases $c$ | DSI | Classification |
|---|---|---|---|
| (a) beam | pin + 2 rollers $\Rightarrow r=4$; one internal hinge $c=1$; $\text{DSI}=r-3-c$ | $4-3-1=0$ | Statically determinate |
| (b) triangular frame | $m=3,\;n=3,\;r=5$ (pin + roller on the top member, pin at the apex), $c=3$ (all three joints pinned) | $3(3)+5-3(3)-3=2$ | Indeterminate, 2° |
| (c) trapezoidal frame | $m=3,\;n=4,\;r=6$ (two fixed bases), $c=2$ (top corners hinged) | $3(3)+6-3(4)-2=1$ | Indeterminate, 1° |
| (d) stepped frame | rigid tree: pin + pin + roller $\Rightarrow r=5$, no internal hinge; $\text{DSI}=r-3$ | $5-3=2$ | Indeterminate, 2° |
| (e) X-braced truss | $m=10,\;n=6,\;r=4$ (two pinned corners); both diagonals present in each of two panels | $10+4-2(6)=2$ | Indeterminate, 2° |
| (f) roof truss | $m=10,\;n=7,\;r=4$ (two wall pins) | $10+4-2(7)=0$ | Statically determinate |
In (a) the roller-under-hinge count leaves exactly three global equations plus the one hinge condition to match the four reactions — determinate. In (e) each panel carries both crossing diagonals (uncoupled where they cross), so the truss is two members redundant. In (f) the second wall pin (a fourth reaction) is exactly offset by the one internal degree of freedom the fan of members would otherwise have, so $m+r-2n=0$ — determinate and stable (a rank check on the equilibrium matrix confirms it is not a mechanism).
