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, read from the examination figure as follows.

Find. For each structure, the classification (unstable / determinate / indeterminate) and, where indeterminate, the degree.

Approach. Count first — for beams and frames $i = r - 3 - c$ (reaction components, less the three equilibrium equations, less one for each condition equation released by an internal hinge or link); for trusses compare $m + r$ with $2j$ — then test the arrangement, because a count of zero only guarantees determinacy if the restraints are neither concurrent nor parallel and the bracing is complete.

1. Part (a) — continuous beam with two hinges. The reaction components are three at the fixed end and one at each of the three rollers, and each internal hinge supplies one condition equation: $$i = r - 3 - c = (3 + 1 + 1 + 1) - 3 - 2 = \boxed{1}$$ The beam is statically indeterminate to the first degree. Every part of it is restrained, so there is no mechanism to look for.
2. Part (b) — two built-in beams joined by a link. Read as drawn, this is two separate rigid bodies. Each carries its own three reaction components at its wall; the link between them transmits one force along its own axis. Counting bodies rather than reactions, $$i = (\text{unknowns}) - (\text{equations}) = (3 + 3 + 1) - (3 \times 2) = \boxed{1}$$ so it is statically indeterminate to the first degree. Each cantilever is stable on its own, so the extra link is a genuine redundancy and not a mechanism.
3. Part (c) — rigid multi-legged frame. There is no closed loop of members and no internal release, so the whole frame is one rigid body and only the external count matters: $$i = r - 3 = (2 + 3 + 2) - 3 = \boxed{4}$$ It is statically indeterminate to the fourth degree. The three support points are not collinear and the reactions are neither all parallel nor concurrent, so it is stable.
4. Part (d) — count says determinate. A single member carries three reaction components, one from each roller, so $$i = r - 3 = 3 - 3 = 0$$ By count the member is statically determinate. That is the trap.
5. Part (d) — the arrangement says otherwise. Put the foot at the origin and the head at $(a,\,a)$, which is what the 45° drawing with the 90° mark on the enclosing triangle means. The foot roller acts vertically, so its line of action is $x = 0$; the head roller bears on the vertical wall and acts horizontally, so its line is $y = a$. These two meet at the corner $C = (0,\,a)$. The mid-length roller acts normal to the member at $M = (a/2,\,a/2)$, and $$(C - M)\cdot \hat{u} = \left(-\tfrac{a}{2},\ \tfrac{a}{2}\right)\cdot \frac{1}{\sqrt2}\,(1,\,1) = \frac{1}{\sqrt2}\left(-\tfrac{a}{2}+\tfrac{a}{2}\right) = 0$$ so the third line of action passes through $C$ as well. All three reactions are concurrent, the member is free to rotate about $C$, and structure (d) is unstable (a mechanism). Note that the concurrency is exact only at 45°: for a head at $(2a,\,a)$ the same dot product is $-0.45a \neq 0$ and the member would be determinate.
6. Part (e) — truss count. Four joints, five members (the two crossing diagonals are separate members, unconnected where they cross), pin plus roller: $$m + r = 5 + 3 = 8 = 2j = 2(4)$$ Building the truss outward — the pinned joint is fixed, the two members meeting at each remaining joint locate it in turn, and the roller takes the last degree of freedom — leaves no free motion, so structure (e) is statically determinate (and stable). Solving the joint equilibrium matrix numerically confirms full rank 8.
7. Part (f) — count says determinate, arrangement says mechanism. Thirteen members, eight joints, three reaction components: $$m + r = 13 + 3 = 16 = 2j = 2(8)$$ But all three supports on the left edge act horizontally — a link at the top and rollers bearing on the vertical wall at the other two. Three parallel reactions cannot equilibrate the two vertical loads: $$\sum F_y = -\,2P \neq 0 \quad \text{for any set of horizontal reactions}$$ Structure (f) is therefore unstable. The equilibrium matrix is rank 15 against 16 equations, one short, which is the algebraic statement of the same fact.