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

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.

Method. For rigid (beam/frame) structures the degree of static indeterminacy is $$DSI = (3m + r) - (3n + c)$$ where $m$ = members, $r$ = reaction components, $n$ = nodes and $c$ = released condition equations (each internal hinge joining two members releases one). Equivalently $DSI = 3(\text{closed loops}) + (r-3) - (\text{releases})$. For pin-jointed trusses, $$DSI = (m + r) - 2n,$$ and the structure is stable only if the equilibrium (statics) matrix has full rank $2n$; a positive $DSI$ with full rank means indeterminate to that degree.

1(a) — Continuous beam, three rollers + fixed end, two internal hinges

Reactions $r = 3(1)+3 = 6$ (three rollers give one vertical each; the built-in end gives $V,H,M$). Condition equations $c = 2$ (two hinges). Treating the collinear beam as one line, $DSI = r-3-c = 6-3-2 = \boxed{1}$. Statically indeterminate, degree 1.

1(b) — Portal frame, two pinned bases, one hinge in the beam–column joint

$m=3$ (two columns + beam), $n=4$, $r = 2(2) = 4$ (two pins), one hinge $c=1$: $$DSI = 3(3)+4-3(4)-1 = 9+4-12-1 = \boxed{0}.$$ Statically determinate. (A rigid two-pin portal is indeterminate to degree 1; the single hinge releases exactly that redundancy.)

1(c) — Two-bay, two-storey frame, three fixed bases, four hinges in the roof

The rigid frame (before hinges) has four closed panels (2 bays × 2 storeys), so $3(4)=12$ degrees, with $r=3(3)=9$: $3m+r-3n = 3(10)+9-3(9)=12$. The four roof hinges release four conditions: $$DSI = 12 - 4 = \boxed{8}.$$ Statically indeterminate, degree 8.

1(d) — Two beams linked by two verticals

$m=8$, $n=8$, $r = 2+1+1+1 = 5$ (upper pin+roller, lower roller+roller), $c=0$: $$DSI = 3(8)+5-3(8) = \boxed{5}.$$ The two verticals close one panel ($+3$) and there are $r-3=2$ external redundants: $3+2=5$. Statically indeterminate, degree 5.

1(e) — Two-diamond truss, pin + roller

$m=14$, $n=8$, $r=3$: $m+r-2n = 14+3-16 = 1$. The $16\times17$ statics matrix has full rank 16, so the truss is stable and indeterminate to degree 1.

1(f) — Two-storey truss on three pinned supports

$m=11$, $n=8$, $r=6$: $m+r-2n = 11+6-16 = 1$. Statics matrix rank 16 (full), so the truss is stable and indeterminate to degree 1 (externally over-supported by two, internally deficient by one).