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Truss Stability with a Wall Roller

Mediumstabilitydeterminacy
An ideal triangular truss has joints A (0, 0), B (4, 0) and C (2, 3), in meters, with members AB, BC and CA. A has a pin support. At B a roller bears against a vertical wall, so it supplies only a horizontal reaction. Using small-displacement first-order analysis, is the support arrangement externally stable, given that m + r = 3 + 3 = 6 = 2j?
  1. AStable and determinate, because m + r = 2j
  2. BStable but indeterminate to the first degree
  3. CUnstable externally, because the reactions cannot resist a moment about A
  4. DUnstable internally, because a truss needs four members
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Answer: C. Unstable externally, because the reactions cannot resist a moment about A

The pin reactions pass through A, and the horizontal reaction at B acts along y = 0, which also passes through A. All three reaction lines are concurrent at A, so none can resist a moment about A: a small rigid rotation about A moves B vertically, which the wall roller allows. The truss is externally unstable even though m + r = 2j. Wrong: the count equality is necessary, not sufficient, for stability; there is no redundancy (so not indeterminate); and three members forming a triangle are internally stable, so a fourth member is not needed.
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