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Trusses are structures where all of the members are connected together at pinned joints. Since each member in a truss has a pin at the end, the members cannot take any moment or shear. The stability and determinacy equations for a truss are simply a special case of the general internal 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} \tag{1} \\ \text{Statically determinate internally: } 3m + r &= 3j + e_c \label{eq:SI-Det} \tag{2} \\ \text{Statically indeterminate internally: } 3m + r &> 3j + e_c \label{eq:SI-Indet} \tag{3} \end{align}

Since there is a hinge at each end of each member, there are $2m$ connections between members and hinged joints in the structure (where $m$ is the number of truss members). To find the number of equations of condition at each joint, the previous equation $e_c = n-1$ applies. This equation tells us that for each hinged joint, there are $e_c = n-1$ equations of condition, where $n$ is the number of members connected to the hinge. This means that each joint has one less equation of condition than the number of members connected to it. Therefore, we must subtract $1$ from the total number of hinged connections at for each joint. The total number of equations of condition for the whole truss then becomes the total number of connections between members connected to hinged joints, minus the number of joints (since we have to subtract $1$ for each joint):

\begin{align} \tag{4} e_c = 2m - j \end{align}

where $m$ is the number of truss members and $j$ is the total number of joints between truss members.

If we sub this value of $e_c$ into the general equation for indeterminacy \eqref{eq:SI-Indet}:

\begin{align*} 3m+r &> 3j + e_c \\ 3m+r &> 3j + (2m-j) \\ m + r &> 2j \end{align*}

If the truss satisfies this inequality, then it is indeterminate. If we repeat this exercise for the other equations \eqref{eq:SI-Unstable} and \eqref{eq:SI-Det}, we get:

\begin{align} \text{Statically unstable internally: } m + r &< 2j \label{eq:Truss-Unstable} \tag{5} \\ \text{Statically determinate internally: } m + r &= 2j \label{eq:Truss-Det} \tag{6} \\ \text{Statically indeterminate internally: } m + r &> 2j \label{eq:Truss-Indet} \tag{7} \end{align}

The degree of indeterminacy for trusses is given by the following equation:

\begin{equation} \tag{8} \boxed{i_e = m + r - 2j \,} \end{equation}

Note that these equations do not necessarily tell the whole story because a truss can still be unstable due to an internal collapse mechanism as will be described in the next section. If you try to solve for the internal forces in such an unstable truss, it will become clear that a solution is not possible.

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