16-Civ-A1 Elementary Structural Analysis · May 2013
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
National Exams, May 2013 — 98-Civ-A1 Elementary Structural Analysis. Three-hour, closed-book examination (approved Sharp/Casio calculator only). Format: six questions constitute a complete paper — answer all of #1–#5 and any one of #6, #7 or #8. All eight questions are solved below for completeness.
Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy, method of joints/sections, virtual-work deflections, moment distribution, slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges and compound structures. Sign convention: sagging bending moment positive; tension member forces positive (T), compression negative (C).
Given. Six planar structures (a–f), with supports, internal hinges and members as drawn on the exam sheet. Beam/frame members transmit axial force, shear and bending; truss members are two-force (axial only).
Find. The stability/determinacy class of each, and the degree of static indeterminacy (DSI) where applicable.
Approach. Count reactions r, members m, joints/nodes n and internal condition releases c (one per internal hinge in a beam/frame). For a beam or rigid frame, $\text{DSI}=(3m+r)-(3n+c)$; for a pin-jointed truss, $\text{DSI}=(m+r)-2n$. A negative value (or a supported rigid-body/partial-collapse mechanism) means unstable; zero means statically determinate; a positive value is the degree of static indeterminacy.
[Figure not reproduced: (a) beam (b) 2-bay frame (c) gable, apex hinge (d) stepped frame (e) truss (f) truss The six structures to classify (redrawn from the exam sheet). See the official exam paper.]
(a) Beam — pin + two rollers + one internal hinge. Reactions $r=4$ (pin 2, two rollers 1 each); one internal hinge gives $c=1$. As a single beam $3m+r-(3n+c)$ reduces to $r-(3+c)=4-(3+1)=0$. $\Rightarrow$ ==**statically determinate and stable**==.
(b) Two-bay portal — fixed / pin / fixed feet. Model as $m=5$ members (two beam spans + three columns), $n=6$ nodes, $r=3+2+3=8$, no internal hinge ($c=0$): $\text{DSI}=3(5)+8-3(6)-0=5$. The members form no closed cell, so this is purely external redundancy ($r-3=5$). $\Rightarrow$ ==**indeterminate to the 5th degree**==.
(c) Trapezoidal (gable) frame — two pinned feet + apex hinge. $m=4$, $n=5$, $r=2+2=4$, apex hinge $c=1$: $\text{DSI}=3(4)+4-3(5)-1=0$. This is the classic three-hinged frame (two support pins + one internal hinge = three hinges). $\Rightarrow$ ==**statically determinate and stable**==.
(d) Stepped frame — fixed / pin / fixed feet. Open (tree-like) rigid frame, $m=5$, $n=6$, $r=8$, $c=0$: $\text{DSI}=3(5)+8-3(6)=5$. Again all external ($r-3=5$, no closed cell). $\Rightarrow$ ==**indeterminate to the 5th degree**==.
(e) Bowstring truss — crossed diagonals, pin + roller. Counting the two crossing (unconnected) diagonals in each half gives $m=11$ two-force members, $n=6$ joints, $r=3$: $\text{DSI}=(m+r)-2n=11+3-12=2$. $\Rightarrow$ ==**indeterminate to the 2nd degree**== (the two redundant crossing diagonals).
(f) Inclined truss — three pinned feet. $m=12$ members, $n=9$ joints, $r=2\times3=6$: $\text{DSI}=(m+r)-2n=12+6-18=0$. The triangulated web plus three pins is just-rigid. $\Rightarrow$ ==**statically determinate and stable**==.
