2.6 Stability

An unstable structure generally cannot be analysed. Therefore, it is useful to know if a structure is stable or unstable before a structural analysis is conducted. There are four main ways that a structure may be geometrically unstable. These apply only to linear geometric stability and not to instability caused by buckling, member yielding or nonlinear geometry.

  1. There are not enough reactions: This will generally be clear from an application of the determinacy equations \eqref{eq:SI-Unstable} to \eqref{eq:SI-Indet}.
    \begin{align}
    \text{Statically unstable internally: } 3m + r &< 3j + e_c \label{eq:SI-Unstable} \\
    \text{Statically determinate internally: } 3m + r &= 3j + e_c \label{eq:SI-Det} \\
    \text{Statically indeterminate internally: } 3m + r &> 3j + e_c \label{eq:SI-Indet}
    \end{align}
  2. The reactions are parallel: All of the reaction components point in the same direction. An example of such a situation is shown in Figure 2.5. In this example, the horizontal equilibrium $\sum {F_x} = 0$ cannot be solved and there will be a net horizontal force on the system with no resistance.
    Figure 2.5: Instability due to Parallel Reactions
  3. The reactions are concurrent: All of the reaction components meet at a point. An example of such a situation is shown in Figure 2.6. Effectively, the system is free to rotate as a rigid body around the point that the reaction components meet at.
    Figure 2.6: Instability due to Concurrent Reactions
  4. There is an internal collapse mechanism: This is any situation in which there is an internal mechanism in the system that will cause it to deform between the supports. In some such situations, this will be clear from the use of the determinacy equations, but in others, it may not. In all such cases, though, the instability will become clear during the structural analysis because it will be impossible to solve for all of the internal forces. An example internal collapse mechanism is shown in Figure 2.7.
    Figure 2.7: Instability due to an Internal Collapse Mechanism

Example

Determine whether the structures shown in Figure 2.8 are externally determinate, internally determinate, externally indeterminate, internally indeterminate or unstable. If a structure is indeterminate, determine how many degrees of indeterminacy it has.

Figure 2.8: Determinacy Examples
    1. External Determinacy:
      \begin{align*}
      i_e &= r - (3+ e_c) \\
      r &= 4, \; e_c = 1
      \end{align*}
      (The hinge on the left at the pin does not provide any additional equations of condition). Therefore, $i_e = 0$. Then, is this structure statically determinate? No, it is unstable because if we take a free-body diagram of the left side of the beam, and take a sum of moments about the center hinge, the sum of moments will be non-zero due to the vertical reaction at the left pin (but we know that it has to be zero due to the existence of the pin).
    2. Internal Determinacy:
      \begin{align*}
      i_e &= (3m+r) - (3j+e_c) \\
      m &= 2, \; r = 4, \; j = 3, \; e_c = 1
      \end{align*}
      (Again, the hinge on the left at the pin does not provide any additional equations of condition). Therefore,
      \begin{align*}
      3m+r = 10, \; 3j+e_c = 10, \text{ and } i_e = 0
      \end{align*}
      Then, is this structure statically determinate? No, it is unstable due to the same reason above.
    1. External Determinacy:
      \begin{align*}
      r = 3, \; e_c = 0
      \end{align*}
      Therefore, $i_e = 0$. Then is this structure statically determinate? No, because the reactions are concurrent through the pin on the right.
    2. Internal Determinacy:
      \begin{align*}
      m = 2, \; r = 3, \; j = 3, \; e_c = 0
      \end{align*}
      Therefore,
      \begin{align*}
      3m+r = 9 \text{ and } 3j+e_c = 9
      \end{align*}
      so the structure appears internally determinate, but it is still unstable due to the concurrent reactions.
    1. External Determinacy:
      \begin{align*}
      r = 3, \; e_c = 0
      \end{align*}
      Therefore, $i_e = 0$. Since there are no sources of instability, this structure is externally statically determinate.
    2. Internal Determinacy:
      \begin{align*}
      m = 6, \; r = 3, \; j = 6, \; e_c = 0
      \end{align*}
      Therefore,
      \begin{align*}
      3m+r = 21 \text{ and } 3j+e_c = 18
      \end{align*}
      so this structure is internally statically indeterminate to three degrees (or "3$^\circ$ S.I.").
    1. External Determinacy:
      \begin{align*}
      r = 5, \; e_c = 2
      \end{align*}
      Therefore, $i_e = 0$. Since there are no sources of instability, this structure is externally statically determinate.
    2. Internal Determinacy:
      \begin{align*}
      m = 5, \; r = 5, \; j = 6, \; e_c = 2
      \end{align*}
      Therefore,
      \begin{align*}
      3m+r = 20 \text{ and } 3j+e_c = 20
      \end{align*}
      so this structure is internally statically determinate (or "S.D.").