Pre-tensioned fasteners

Consider the figure below, which shows a bolt under pre-tension.

initial fastener forces

The bolt is pre-loaded to a tension T_b, and the plate(s) is pre-loaded to a compression C_i.

The lengthening of the bolt under this pre-tension is:

    \begin{equation*} (\delta_i)_{bolt} = \frac{T_b L_b}{A_b E_b} \end{equation*}

The shortening of the plate (or in many cases, multiple plates) under this pre-compression is:

    \begin{equation*} (\delta_i)_{plate}=\frac{C_i t_p}{A_p E_p} \end{equation*}

The figure below shows the loads on the bolt and plate after an external force is applied. In the figure, the example that is shown is a moment connection that uses angles instead of welds. Under application of a beam moment, each leg of each angle is pulled in a way that causes some additional tension in the bolt and some reduction in compression of the plate. This force that’s applied to the connection assembly is denoted P in the figure. What we’ll prove is that only a small fraction of this force, P, causes additional tension in the bolt, so long as the bolt has been sufficiently pre-tensioned.

final fastener forces

First, let’s look at the additional lengthening of the bolt, in terms of the force T_f, which is so far unknown.

(\delta_f)_{bolt} = (\delta_i)_{bolt} + \delta_{add} = \frac{T_b L_b}{A_b E_b} + \delta_{add}

\rightarrow \delta_{add}=(\delta_f)_{bolt} - \frac{T_b L_b}{A_b E_b}

(1)   \begin{equation*} \delta_{add}=\frac{T_f L_b}{A_b E_b} - \frac{T_b L_b}{A_b Eb} \end{equation*}

Now, let’s look at the lengthening of the plate(s), which is written in terms of unknown variables C_i and C_f.

(\delta_f)_{plate}=(\delta_i)_{plate} - \delta_{decrease}=\frac{C_i t_p}{A_p E_p} -\delta_{decrease}

\rightarrow \delta_{decrease} = \frac{C_i t_p}{A_p E_p} - (\delta_f)_{plate}

(2)   \begin{equation*} \delta_{decrease}=\frac{C_i t_p}{A_p E_p} - \frac{C_f t_p}{A_p E_p} \end{equation*}

We’d like to find an expression that relates T_f to T_b and P. In other words, we’d like to know how the pre-load in the bolt, T_b, affects the final tension in the bolt, T_f, when the connection assembly is subjected to external tensile loads, P. We currently have two equations and many “unknowns” (\delta_{add}, \delta_{decrease}, C_i, C_f, T_f). We need to consider force equilibrium and displacement compatibility in order to solve.

Since the bolts and the plates “lengthen” by the same amount:

(3)   \begin{equation*} \delta_{add} = \delta_{decrease} \end{equation*}

From equilibrium of our two free-body diagrams (prev. figures that contain T_b, C_i, and T_f, C_f, P), we have two additional equations:

(4)   \begin{equation*} C_i = T_b \end{equation*}

(5)   \begin{equation*} P+C_f=T_f \end{equation*}

bolt length and plate thickness

We can see in the above figure that:

    \begin{equation*} L_b = t_p \end{equation*}

Additionally, we can assume that the bolt and the plate have the same E (they are both steel), so that:

    \begin{equation*} E_b = E_p \end{equation*}

We now have 5 equations (1 2 3 4 5) and 5 unknowns (\delta_{add}, \delta_{decrease}, C_i, C_f, T_f), so we can solve for the desired result (T_f expressed in terms of T_b and P):

(6)   \begin{equation*} T_f = T_b + \frac{P}{1+ A_p/A_b} \end{equation*}

From eq. 6, we can conclude that only a small fraction of the load “P” is actually “added” to the bolt, so long as a) the plate is big and b) the bolt has been sufficiently pre-tensioned so that the bolt and plate remain in contact when “P” is applied. Eq. 6 would not be applicable if the bolt is not sufficiently pre-tensioned (e.x. T_f would simply be equal to P, if T_b is zero).

Pre-tensioned bolts have a relatively constant tension, even if the external loading (i.e.P”) is cyclical. This relatively constant tension is necessary to avoid bolt failure due to fatigue.