Study log

Engineers Notebook

Stability and Determinacy of Structures

Theory of Structures
Nelson
Latest Update: 2025-11-12 09:11:45
  1. Determinacy relates to the analysis of how fully a structure's internal forces and reactions can be determined using equations of equilibrium. A structure is considered determinate if all the internal forces and support reactions can be precisely calculated using static equilibrium equations.
  2. Stability refers to the ability to maintain equilibrium and resist failure or collapse. A stable structure is able to maintain equilibrium when subjected to external loads or forces, without undergoing excessive deformation or displacement. Statically Unstable occurs when the state of balance is disrupted, typically due to an inadequate number of supporting elements within a structure.
Instability or movement of a structure or a member can develop because of improper constraining by the supports. This can happen if the support reaction are concurrent (Figure 1) at a point or reactions are parallel (Figure 2)enggportalenggportal

Formulations for Stability and Determinacy of Beams, Truss and Frames

Structure Type Condition Stability / Determinacy
Beam r < 3 + c Beam is unstable
r = 3 + c Beam is stable and determinate
r > 3 + c Beam is stable and indeterminate
Truss b + r < 2j Truss is unstable
b + r = 2j Truss is stable and determinate
b + r > 2j Truss is stable and indeterminate
Frame 3b + r < 3j + c Frame is unstable
3b + r = 3j + c Frame is stable and determinate
3b + r > 3j + c Frame is stable and indeterminate

Where:
r = Total number of reactions
c = Equations of condition (c = 1 for a hinge; c = 2 for a roller; c = 0 for a beam with internal pin connection)
b = Number of members
j = Number of joints

Beam Stability and Determinacy Examples

Beam 1
Given Computation Condition Result
R = 5, m = 1 5 > 3(1) r > 3 + c Indeterminate
R = 5, C = 0 5 > 3 + 0 r > 3 + c Beam is stable and indeterminate
Beam 2
Given Computation Condition Result
R = 7, m = 2 7 > 3(2) r > 3 + c Indeterminate
R = 5, C = 1 5 > 3 + 1 r > 3 + c Beam is stable and indeterminate
Beam 3
Given Computation Condition Result
R = 6, m = 2 6 = 3(2) r = 3 + c Determinate
R = 4, C = 1 4 = 3 + 1 r = 3 + c Stable and determinate
Beam 4
Given Computation Condition Result
R = 11, m = 4 11 < 3(4) r < 3 + c Unstable
R = 7, C = 2 7 > 3 + 2 r > 3 + c Unstable internally
Beam 5
Given Computation Condition Result
R = 7, m = 4 7 < 3(4) r < 3 + c Unstable
R = 4, c = 3 4 < 6 r < 3 + c Unstable

More on Theory of Structures