16-Civ-A1 Elementary Structural Analysis · December 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, December 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; upward reactions positive; tension member forces positive (T), compression negative (C).
Approach. Count external reaction components r, released internal conditions c (each internal hinge in a beam releases one moment; a pin joining k members releases k−1), and apply the appropriate degree-of-static-indeterminacy (DSI) formula: beams/frames DSI = 3m + r − 3n − c; pin-jointed trusses DSI = m + r − 2n (where m = members, n = joints). DSI > 0 ⇒ indeterminate to that degree; DSI = 0 ⇒ determinate; a negative count or a geometric defect ⇒ unstable.
(a) Continuous beam, three rollers + fixed end, two internal hinges. Reactions $r = 3(1)+3 = 6$; conditions $c = 2$ hinges. Treating the run of collinear beam members as one flexural chain, $DSI = r-3-c = 6-3-2 = \boxed{1}$ — indeterminate, degree 1.
(b) Continuous beam, two pins + two rollers, one internal hinge. $r = 2+2+1+1 = 6$, $c = 1$. $DSI = 6-3-1 = \boxed{2}$ — indeterminate, degree 2 (the extra horizontal pin restraint is one of the two redundancies).
(c) Single-bay, two-storey frame; both beams pin-connected (hinged) to continuous columns fixed at their bases. Model $m = 6$ members, $n = 6$ joints, $r = 3+3 = 6$ (two fixed feet), $c = 4$ (a hinge at each of the four beam ends). $DSI = 3(6)+6-3(6)-4 = \boxed{2}$. Check by closed loops: two rigid ring-circuits close through the foundation, $2\times3 = 6$ redundancies, less the 4 releases $= 2$. Indeterminate, degree 2.
(d) Gabled frame: two rafters meeting at a hinged apex, a tie beam, three columns fixed at their bases; hinges at both eaves. $m = 7$, $n = 7$, $r = 3(3) = 9$ (three fixed feet), $c = 5$ (two-member release of 1 at the apex, and a three-member pin release of 2 at each eave). $DSI = 3(7)+9-3(7)-5 = \boxed{4}$ — indeterminate, degree 4.
(e) Four-panel parallel truss; the two end panels carry crossed (double) diagonals, pin at one foot and roller at the other. $m = 19$, $r = 3$, $n = 10$. $DSI = m+r-2n = 19+3-20 = \boxed{2}$ — indeterminate, degree 2 (one redundant diagonal in each X-panel).
(f) Rotated-square (diamond) truss on two legs, pin + roller feet, one pair of crossed diagonals. $m = 10$, $r = 3$, $n = 6$. $DSI = 10+3-12 = \boxed{1}$ — indeterminate, degree 1 (the square panel carries one redundant diagonal).
Structure
Classification
(a) beam, 2 hinges
Statically indeterminate — degree 1
(b) beam, 1 hinge
Statically indeterminate — degree 2
(c) 2-storey pinned-beam frame
Statically indeterminate — degree 2
(d) hinged gable + tie, 3 fixed feet
Statically indeterminate — degree 4
(e) 4-panel truss, 2 X-panels
Statically indeterminate — degree 2
(f) diamond truss
Statically indeterminate — degree 1
Check (figure interpretation): the released-condition count for (c) and (d) depends on reading each drawn circle as a true frictionless pin connecting all members meeting at that joint. If a joint circle is instead a single-member release, the degrees fall to (c) 2 and (d) 6/5; the classifications (indeterminate) are unchanged. All base supports in (d) are fixed.