Learn how to analyse bolts in a bolt group for in-plane and out-of-plane loading using the ICR concept and check shear, tension, and combined forces, according to AS 4100.
Introduction
Connections in building structures use a minimum of two bolts and often more than eight bolts. The bolts used in a connection form a bolt group. A bolt group may be acted on by loads and bending moments in the plane of the bolt group (in-plane) or at right angles to it (out-of-plane).
A typical connection with a bolt group is a beam to column bolted connection.
\( \bf{1}\) Bolt Capacity
Bolts are required to be checked individually for shear and tensile loads, as well as with a combined action check.
The shear and tensile capacities of bolts are both based on the minimum tensile strength \( f_{uf} \) rather than the yield strength.
\( \bf{1.1}\) Bolts in shear
As per Cl 9.2.2.1 AS4100, the shear capacity of a bolt \( \phi V_f \) is given by:
$$ \phi V_f = \phi \times 0.62 \times k_r \times f_{uf} (n_n A_c + n_x A_o) $$
Where:
\(\rlap{\phi}\hspace{2.5em}\): capacity reduction factor and is always 0.8, as per Table 3.4 of AS 4100
\(\rlap{k_r}\hspace{2.5em}\): reduction factor for bolted splice connections
\( \rlap{f_{uf}}\hspace{2.5em}\): minimum tensile strength of the bolt
\(\rlap{A_c}\hspace{2.5em}\): core area (at the root of the threads)
\(\rlap{A_o}\hspace{2.5em}\): shank area of the bolt
\(\rlap{n_n}\hspace{2.5em}\): number of shear planes in the threaded regions
\(\rlap{n_x}\hspace{2.5em}\): number of shear planes in the unthreaded region
👉A bolt will either be classed with "threads included" which is standard practice, or "threads excluded" which is non-standard practice. Therefore the \( n_n A_c\) or \( n_x A_o \) will become zero in the equation above, respectively.
As per Cl 9.2.3.1 of AS4100, an additional serviceability limit state check must be performed for friction-type connections (i.e. for the /TF bolting category) where connection slip is intended to be prevented at serviceability loads. The shear capacity of a bolt for a friction-type connection, \( \phi V_{sf} \) is:
$$ \phi V_{sf} = \phi \times \mu n_{ei} N_{ti} k_h$$
Where:
\( \rlap{\phi}\hspace{2.5em}\): capacity reduction factor and is 0.7 in this "special" serviceability check as per Cl 3.5.5 of AS4100
\(\rlap{k_h}\hspace{2.5em}\): factor for hole type, taken as 1.0 for standard holes, 0.85 for oversize holes and short slots, and 0.70 for long slotted holes
\( \rlap{\mu}\hspace{2.5em}\): slip factor, which is the coefficient of friction between plies and depends on the surface preparation of 8.8/TF bolts, varying from 0.05 to 0.35
\( \rlap{N_{ti}}\hspace{2.5em}\): minimum bolt tension imparted to the bolts during installation, and is typically tabulated per bolt size
\(\rlap{n_{ei}}\hspace{2.5em}\): number of shear planes
\( \bf{1.2}\) Bolts in tension
As per Cl 9.2.2.2 AS4100, the tension capacity of a bolt \( \phi N_{tf} \) is given by:
$$ \phi N_{tf} = \phi A_s f_{uf} $$
Where:
\( \rlap{A_s}\hspace{2.5em}\): tensile stress area of the bolt
\( \rlap{f_{uf}}\hspace{2.5em}\): minimum tensile strength of the bolt
\( \bf{1.3}\) Bolts in combined shear and tension
As per Cl 9.2.2.3 AS4100, the combined shear and tension bolt check is given by:
$$ \left(\frac{V^*_f}{\phi V_f}\right)^2 + \left(\frac{N^*_{tf}}{\phi N_{tf}}\right)^2 \leq 1.0 $$
\( \bf{1.4}\) Tabulated design capacities
It is common for textbooks to tabulate bolt capacities per bolt size. See below example tables from Australian Guidebook for Structural Engineers and the Steel Designers' Handbook that provide the tensile and shear (threaded and non-threaded) ULS capacities per bolt size for 4.6/S, 8.8/S, 8.8/TB and 8.8/TF, based on a single shear plane.
\( \bf{2}\) Analysis of Bolt Groups
Bolt groups are subjected to in-plane and out-of-plane loading. Loads on individual bolts are calculated by using a bolt group analysis. In summary, the analysis of bolt groups follows these steps:
- The centroid of the bolt group is evaluated based on the inputted bolt group geometry.
- All applied loads \((F^*_x, F^*_y, M^*_z, V^*_o)\) are calculated as a concentrated resultant load \((F^*_x, F^*_y, M^*_1, M^*_o)\) at the centroid of the bolt group.
- The resultant loads are distributed to each bolt by calculating the shear force \(V^*_f\) and tension force \(N^*_{tf}\) in each bolt, which is proportional to the distance from the bolt to the group centroid.
- The 'critical' bolt is considered to be the bolt furthest from the centroid, which is used for the design check on the overall bolt group.
Let's look at how the in-plane and out-of-plane loading is distributed to each bolt in more detail below.
\( \bf{2.1}\) In-plane loading
As per Cl 9.3 of AS4100, elastic analysis of weld groups for in-plane loading is permitted using the Instantaneous Centre of Rotation (ICR) concept. The ICR is the point at which the bolt group rotates about when subject to overall bolt group actions. The ICR enables us to calculate the distribution of loads to each bolt in a bolt group.
The method is not described further in AS4100, but is summarised below based on guidance in Steel Designers Handbook.
Analysis of the bolt group uses the ICR concept together with superposition. For a bolt group with in-plane design loading, a pure moment acting on a bolt group has the ICR positioned at the bolt group centroid. Whereas, when the same bolt group is subject to shear force only, the ICR is at infinity. Therefore, for bolt group seeing in-plane shear and moments, superposition of the two individual action effects means uniformly distributing shear forces to all bolts in the group while also assuming the bolt group rotation from moment effects occurs about the group centroid.
👉 Based on superposition of in-plane loading, the bolt group ICR is in the same position as the bolt group centroid.
Design actions \( (F^*_x, F^*_y, M^*_z)\) applied away from the centroid of the weld group may be treated as being applied at the centroid plus moments, with forces \( ( F^*_x, F^*_y)\) and a resolved moment \(M^*_1\).
The in-plane design force per bolt,\( V^*_f \) is:
$$ \begin{align*}
V^*_f &= \sqrt{[V^*_x]^2 + [V^*_y]^2} \\
&\text{for:} \\
V^*_x &= \frac{F^*_x}{n} - \frac{M^*_1 y_n}{I_p} \\
V^*_y &= \frac{F^*_y}{n} + \frac{M^*_1 x_n}{I_p} \\
M^*_1 &= F^*_x e_y + F^*_y e_x - M^*_z
\end{align*} $$
Where:
\( \rlap{V^*_x, V^*_y}\hspace{3em}\): x- & y-axis design forces in each bolt
\( \rlap{x_n,y_n }\hspace{3em}\): horizontal and vertical distances, respectively, from bolt to bolt group centroid
\( \rlap{M^*_o}\hspace{3em}\): resolved in-plane moment about the group centroid
\( \rlap{n}\hspace{3em}\): total number of bolts in the bolt group
\( \rlap{I_p}\hspace{3em}\): polar second moment of area of the bolt group
The bolt group centroid coordinates \( \bar{x}, \space \bar{y}\) are given by:
$$\begin{align*}
\bar{x} &= \frac{\sum x_i}{n} \\
\bar{y} &= \frac{\sum y_i}{n}
\end{align*}$$
Where:
\( \rlap{x_i, y_i}\hspace{2.5em}\): coordinates of the bolts
\( \rlap{n}\hspace{2.5em}\): total number of bolts in the bolt group
\( \bf{2.2}\) Out-of-plane loading
Out-of-plane loading is ultimately axial loading on bolt groups. A shear force \( V^*_o\) applied out-of-plane to the bolt group at an eccentricity e, results in a moment \( M^*_o\) which then induced axial forces in the bolts.
AS 4100 does not describe how to determine how much axial load is in each bolt. We have summarised a method below based on guidance in Steel Designers Handbook.
From force/moment equilibrium principles, there are bolts which are not loaded since they are positioned in the bearing (compression) part of the connection. The bolts in the tension region have tension loads that can be evaluated by assuming a linear distribution of force from the neutral axis to the farthest bolts, as shown in the image above. However it is difficult to accurately determine where the neutral axis (NA) exists due to the bolt, plate and support flexibility. A conservative approach, adopted by this calculator, is to assume the NA is at the bolt group centroid line.
From equilibrium principles and the principle of proportioning from similar triangles, the out-of-plane design tension force per bolt, \( N^*_{tf}\) is:
$$N^*_{tf,i} = \frac{M^*_o y_i}{\sum[y_i(y_i + y_c)]} \frac{1}{n_{col}} $$
Where:
\( \rlap{M^*_o}\hspace{2.5em}\): resolved out-of-plane moment about the bolt group centroid, taken as \(M^*_o=V^*_oe\)
\( \rlap{y_i }\hspace{2.5em}\): vertical distance of a bolt to the NA
\( \rlap{y_c}\hspace{2.5em}\): distance from the NA to the compression force, which we conservatively assume is the y-coordinate of the bolt group centroid
\( \rlap{n_{col}}\hspace{2.5em}\): number of columns in the bolt group
