# 2.3 External Indeterminacy

If a structure is *externally determinate*, then all of the reactions may be calculated using equilibrium alone. To calculate external determinacy, the following equations are used:

\begin{align}

\text{Statically unstable externally: } r < 3 + e_c \label{eq:SE-Unstable} \\

\text{Statically determinate externally: } r = 3 + e_c \label{eq:SE-Det} \\

\text{Statically indeterminate externally: } r > 3 + e_c \label{eq:SE-Indet}

\end{align}

where $r$ is the number of reaction components, and $e_c$ is the number of equations of condition. Both of these are described in detail below.

The degree of indeterminacy is given by the following equation:

\begin{equation}

\boxed{i_e = r - (3 + e_c)} \label{eq:SE-DegreeIndet}

\end{equation}

## Reaction Components

In the equations above, $r$ is equal to the total number of reaction components as shown in Table 2.1.

For multiple reaction points, $r$ is the sum of all the components for all the reaction points in the structure.

## Equations of Condition

Additionally, $e_c$ is the number of *equations of condition*. These are release conditions within the structure that provide extra equilibrium equations beyond the three for global equilibrium.For example, if an internal hinge is added to the structure, as shown in Figure 2.2, then there is one equation of condition. If there was no internal hinge in this example, then the structure would be indeterminate and it would not be possible to find the reaction forces or the internal forces (since it has four reaction components). The addition of the hinge provides an additional equilibrium condition which forces the internal moment to be equal to 0 at point B ($\sum {M_B} = 0$). This may be seen if the structure is split into two free body diagrams as shown in the lower part of Figure 2.2. At point B, there are three internal force components that exist in equal and opposite action/reaction pairs on either side of point B:

- Axial Force: $B_x^{AB}$ and $B_x^{BC}$
- Shear Force: $B_y^{AB}$ and $B_y^{BC}$
- Moment: $M_B^{AB}$ and $M_B^{BC}$

So, $M_B^{AB} = M_B^{BC} = 0$, because they are action reaction pairs.

Therefore, only one extra equilibrium equation is possible due to the introduction of the hinge: either $\sum {M_B^{AB}} = 0$ or $\sum {M_B^{BC}} = 0$ but not both because the two equations are not independent. So, for each internal hinge in a structure, there is a single equation of condition: $e_c = 1$.

For a structure with an internal roller, such as that shown in Figure 2.3, both the force transfer in the direction of the roller and the moment are equal to zero at the location of the roller. This provides two extra equilibrium equations, and therefore two equations of condition. For the structure shown in Figure 2.3, the extra equations are:

\begin{align*}

\sum {M_B^{AB}} &= 0 \text{ or } \sum {M_B^{BC}} = 0 \\

\text{and} \\

\sum {B_x^{AB}} &= 0 \text{ or } \sum {B_x^{BC}} = 0

\end{align*}

So, for each internal roller, there are two equations of condition: $e_c = 2$.

If there are additional members that frame into a single internal hinge, then there is an additional equation of condition for each additional member. For example, for three members connected at a hinge, then there are two extra independent equilibrium equations that are added to the system (because in an equal and opposite action/reaction pair, there can only be two sides). So, for a hinge connection with multiple elements, $e_c = n - 1$ where $n$ is equal to the number of members connected to the hinge. Similarly, for a roller connection with multiple members, each additional member adds two equations of condition, $e_c = 2 * (n - 1).$

In summary:

\begin{align}

\text{For a Hinge: }e_c &= n - 1 \label{eq:ec-hinge} \\

\text{For a Roller: }e_c &= 2 (n - 1)

\end{align}

where $n$ is the number of members connected to the hinge or roller.