Question 1 of 8: Stability and Degree of Static Indeterminacy

Question 1: Stability and Degree of Static Indeterminacy (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 plane structures, redrawn below exactly as they appear on the examination paper.

Find. For each structure, its classification — unstable, statically determinate, or statically indeterminate with the degree stated.

[Figure not reproduced: The six structures of Question 1, redrawn from the examination figure. (a)–(d) are beam and frame assemblies; (e) and (f) are pin-jointed trusses. See the official exam paper.]

Approach. Count constraints against equations: for beam and frame assemblies compare the total number of unknown reaction and internal force components with the number of available equilibrium and condition equations, and for the trusses use $m + r$ against $2j$ — then, in every case, partition the structure and ask what motion is left free, because a count alone can never prove stability.

1. Fix the two counting rules before touching any figure. For a beam or frame made of rigid pieces the degree of static indeterminacy is $$i = r - (3 + c)$$ where $r$ is the number of reaction components, $3$ is the number of equilibrium equations for the whole assembly and $c$ is the number of condition (release) equations that internal hinges supply. For a pin-jointed truss with $m$ members, $j$ joints and $r$ reaction components, $$i = (m + r) - 2j .$$ Both formulas are necessary but not sufficient: $i = 0$ with a bad arrangement still gives a mechanism, so each result is checked by inspection afterwards.
2. (a) Propped cantilever with three rollers and one hinge. The left end runs into hatching, so it is fixed and contributes three components; the three triangles on rollers contribute one each, giving $r = 3 + 1 + 1 + 1 = 6$. One internal hinge supplies one condition equation, so $3 + c = 4$ and $$i = 6 - 4 = \boxed{2}$$ The beam is statically indeterminate to the second degree. It cannot be a mechanism: the fixed end alone would hold the beam, and every roller adds a further restraint.
3. (b) Two beams built into one wall, the upper bearing on the lower. The hatched wall on the left runs the full height of the figure, so both beams are built into it; the upper beam then ends on a roller that rides on the lower beam, and the lower beam ends on a roller on the ground. Counting the assembly as two rigid bodies gives $3 + 3 = 6$ equilibrium equations against $3 + 3 + 1 = 7$ external components plus the one interface force the intermediate roller transmits, so $$i = 8 - 6 = \boxed{2}$$ and the assembly is statically indeterminate to the second degree. Each beam separately is already a propped cantilever, and the shared roller couples them.
4. (c) Stepped three-column frame on three pins. Take the joints as: the base of each column, the two beam-to-column corners and the point on the middle column where the lower beam frames in. That is $j = 7$ joints joined by $m = 6$ rigid members (three column lengths counted as the upper and lower parts of the middle column, plus two beams), with $r = 2 + 2 + 2 = 6$. The member-based count is $$i = 3m + r - 3j = 18 + 6 - 21 = \boxed{3}$$ Cross-checking by the loop rule gives the same answer: the frame plus the ground closes two loops, worth $2 \times 3 = 6$ redundants, and the three pins release three moments, so $6 - 3 = 3$. The frame is statically indeterminate to the third degree.
5. (d) Two-storey, two-bay frame on three pins. Now $m = 8$ (each outer column split at the intermediate beam level, the two beams each split at the middle column, and the middle column itself) and $j = 8$, with $r = 6$, so $$i = 3(8) + 6 - 3(8) = \boxed{6}$$ The loop check again agrees: three closed loops give nine redundants and the three pin bases release three moments, leaving six. The frame is statically indeterminate to the sixth degree.
6. (e) Symmetric truss on a pin and a roller. Count members carefully, because the diamond of diagonals is easy to under-count: two rafter segments each side (4), two bottom-chord segments (2), the horizontal middle chord in two segments (2), the vertical from the apex to the centre joint (1), and the two diagonals rising from the bottom centre joint (2) — so $m = 11$. There are $j = 7$ joints and $r = 2 + 1 = 3$, giving $$i = (11 + 3) - 2(7) = \boxed{0}$$ The count says determinate, and the arrangement confirms it. The rank of the $14 \times 14$ joint-equilibrium matrix built from the geometry is a full $14$, so the truss is statically determinate and stable. (The apex-to-centre vertical is a zero-force member under the loading shown, which is not the same thing as being redundant — it is what stops the centre joint from moving vertically.)
7. (f) Three-legged truss on three pins. Follow each line from the apex to the ground: the apex is joined to the three mid-height chord joints, two of those lines continue straight on to the left and middle pins, the third runs vertically to the right pin, the mid-height chord contributes two segments, and one further diagonal crosses from the left chord joint to the right pin without connecting where it crosses. That gives $m = 9$ members and $j = 7$ joints, with $r = 3 \times 2 = 6$, so $$i = (9 + 6) - 2(7) = \boxed{1}$$ The equilibrium matrix has rank $14$, equal to the number of equations, so every load can be carried and the assembly is stable. The truss is statically indeterminate to the first degree.