In steel connection design, bolts are usually designed as a bolt group that acts as one body to resist a given load. The strength of a bolt group is normally set by the controlling strength of its most critical bolt. The direct loads are distributed among the total number of bolts, while the induced moment due to the load eccentricity is distributed in relation to the bolt group’s moment of inertia and distance from the centroid. This is called elastic analysis. Because of its simplified and conservative assumptions about load distribution, it often yields over-designed bolted connections.
When talking about value engineering and economical designs, the inelastic approach is preferred by most fabricators. It requires fewer bolts for the same magnitude of load. To do the inelastic approach, the instantaneous centre of rotation (ICOR) method using iterations is the best way.
In this article, we demonstrate how to calculate the strength of a bolted connection using the ICOR method. The reactions per bolt are calculated using Equation (7-1) on page 7-7 of the AISC 15th Edition Manual. That is then used to check whether the assumed location of the instantaneous centre of the bolt group is correct. Finally, once we have the correct IC location, we calculate the bolt group coefficient C to determine its strength.
Worked by hand, the ICOR method is a long process, because finding the Instantaneous Centre (IC) is trial and error. With a computer solver the IC of a bolt group is found by programmed iterations instead. The SkyCiv Bolt Group Solver uses a fast iteration method to determine the IC location and the bolt group coefficient in seconds. It is used by both the AISC and AS 4100 connection modules.
The worked example below is the same calculation the solver performs, shown long hand so you can follow what it is doing.
Getting the Bolt Group Properties
Let’s start with a simple analysis of a bolt group of four bolts loaded with an eccentric vertical shear load of 10 kips. The eccentricity of the load along the x axis is 4 inches to the right of the bolt group. The angle from the vertical is zero and the eccentricity along the y axis is zero.
[math] V_{u} = 10\,kips [math]
[math] \theta = 0\,deg [math]
[math] e_{x} = 4\,in [math]
[math] e_{y} = 0\,in [math]
The first thing to do is to get the coordinates of all the bolts in the group. Visual guides and tables are highly recommended.
| Bolt ID | X (in) | Y (in) |
| 1 | 0 | 0 |
| 2 | 0 | 3 |
| 3 | 3 | 0 |
| 4 | 3 | 3 |
To get the centroid of the bolt group along the x and y axes, we need the formula below.
Let [mathin] n [mathin] = total number of bolts
[math] X_{CG} = \frac{\sum X}{n} [math]
[math] Y_{CG} = \frac{\sum Y}{n} [math]
Then, our solution is:
[math] X_{CG} = \frac{\sum X}{n} = \frac{0\,in + 0\,in + 3\,in + 3\,in}{4} = 1.5\,in [math]
[math] Y_{CG} = \frac{\sum Y}{n} = \frac{0\,in + 3\,in + 0\,in + 3\,in}{4} = 1.5\,in [math]
Assume the location of the I.C.
After getting the centroid, we assume the location of the instantaneous centre (IC). As a first try, we can assume the IC sits at the geometric centroid of the bolt group.
So, assume
[math] X_{IC} = X_{CG} = 1.5\,in [math]
[math] Y_{IC} = Y_{CG} = 1.5\,in [math]
Then we tabulate the displacement of each bolt from the IC. Get the distance along x and the distance along y first, then the resultant displacement.
| Bolt ID | cx (in) | cy (in) | c (in) |
| 1 | -1.5 | -1.5 | 2.121 |
| 2 | -1.5 | 1.5 | 2.121 |
| 3 | 1.5 | -1.5 | 2.121 |
| 4 | 1.5 | 1.5 | 2.121 |
Where,
[math] c_{x} = X_{i} – X_{IC} [math]
[math] c_{y} = Y_{i} – Y_{IC} [math]
[math] c = \sqrt{ {\left( c_{x} \right)}^{2} + {\left( c_{y} \right)}^{2} } [math]
For Bolt No. 1, our solution is
[math] c_{x} = 0\,in – 1.5\,in = -1.5\,in [math]
[math] c_{y} = 0\,in – 1.5\,in = -1.5\,in [math]
[math] c = \sqrt{ {\left( -1.5\,in \right)}^{2} + {\left( -1.5\,in \right)}^{2} } = 2.121\,in [math]
Calculate the deformation per bolt wrt distance from IC
After getting the bolt distances from the assumed IC location, we calculate the deformation of each bolt as a function of its distance.
The maximum deformation per bolt, [mathin] \Delta_{max} = 0.34\,in [mathin], is based on experimental data for an ASTM bolt as described in AISC page 7-8. By linear proportion, and setting [mathin] \Delta_{max} = 0.34\,in [mathin], we can calculate the deformation of an individual bolt relative to its share of the maximum distance [mathin] c_{max} [mathin]. The equation for getting [mathin] \Delta [mathin] is shown below.
[math] \Delta_{1} = 0.34\,in \times \left( \frac{c}{c_{max}} \right) [math]
For Bolt No. 1, the deformation is
[math] \Delta_{1} = 0.34\,in \times \left( \frac{2.121\,in}{2.121\,in} \right) = 0.34\,in [math]
Because the IC was assumed at the centroid, all four bolts are the same distance from it, so all four reach the same deformation. The calculated deformations are tabulated below.
| Bolt ID | [mathin] \Delta [mathin] (in) |
| 1 | 0.34 |
| 2 | 0.34 |
| 3 | 0.34 |
| 4 | 0.34 |
Get the reactions per bolt
Once we have the deformation per bolt, we use AISC 15th Ed. Eq (7-1) to get the reactions per bolt.
[math] R = R_{ult} \left( 1 – e^{-10\Delta} \right)^{0.55} [math]
The [mathin] R_{ult} [mathin] in the equation is the assumed ultimate load on a bolt, which we can set as the bolt shear strength.
[math] R_{ult} = \phi R_{n} [math]
For our example, we use a bolt shear strength of [mathin] 24.4\,kip [mathin]. Another value is equally valid, because it cancels out when we calculate the bolt group coefficient [mathin] C [mathin] later on.
For Bolt No. 1, the calculated reaction is
[math] R = R_{ult} \left( 1 – e^{-10\Delta} \right)^{0.55} [math]
[math] R = 24.4\,kip \left( 1 – e^{-10 \times \left( 0.34\,in \right)} \right)^{0.55} [math]
[math] R = 23.949\,kip [math]
For the rest of the bolts, the calculated reactions are as follows. The components of the bolt reaction [mathin] R [mathin] along x and y are also shown.
| Bolt ID | R (kip) | Rx (kip) | Ry (kip) |
| 1 | 23.949 | 16.937 | -16.937 |
| 2 | 23.949 | -16.937 | -16.937 |
| 3 | 23.949 | 16.937 | 16.937 |
| 4 | 23.949 | -16.937 | 16.937 |
| ⅀Rx = 0 | ⅀Ry = 0 |
Each bolt force acts perpendicular to the line joining that bolt to the IC, which is what sets the sign of its two components. For Bolt No.1, the solutions for the x and y components are shown below.
[math] R_{x} = -R \left( \frac{c_{y}}{c} \right) = -23.949 \times \left( \frac{-1.5\,in}{2.121\,in} \right) = 16.937\,kip [math]
[math] R_{y} = R \left( \frac{c_{x}}{c} \right) = 23.949 \times \left( \frac{-1.5\,in}{2.121\,in} \right) = -16.937\,kip [math]
Next we need the moment each bolt reaction produces about the IC. Use the components [mathin] R_{x} [mathin] and [mathin] R_{y} [mathin] with the lever arms [mathin] c_{y} [mathin] and [mathin] c_{x} [mathin].
[math] M_{r} = -R_{x} c_{y} + R_{y} c_{x} [math]
For Bolt No.1, the moment reaction about the IC is
[math] M_{r} = -16.937\,kip \times \left( -1.5\,in \right) + \left( -16.937\,kip \right) \times \left( -1.5\,in \right) [math]
[math] M_{r} = 50.811\,kip\text{-}in [math]
As a check, the same moment is [mathin] R \times c = 23.949 \times 2.121 = 50.8\,kip\text{-}in [mathin], because the force is perpendicular to the radius.
Every bolt here sits the same distance from the assumed IC and carries the same force, so every bolt returns the same moment. The moment reactions are tabulated below.
| Bolt ID | Mr (kip-in) |
| 1 | 50.811 |
| 2 | 50.811 |
| 3 | 50.811 |
| 4 | 50.811 |
| ⅀Mr = 203.244 |
Verifying the IC location
Now that we have the shear and moment reactions per bolt, we use them to determine the load this bolt group resists. Take the resultant of the sum of all reactions along x and the sum of all reactions along y.
From the previous section, we calculated that
[math] \sum R_{x} = 0\,kip [math]
and
[math] \sum R_{y} = 0\,kip [math]
So,
[math] P_{u} = \sqrt{ {\left( \sum R_{x} \right)}^{2} + {\left( \sum R_{y} \right)}^{2} } = 0\,kip [math]
The resulting load is [mathin] P_{u} = 0\,kip [mathin]. A bolt group rotating about its own centroid carries pure moment and no net force, so it cannot resist the applied shear. That already tells us the first assumed IC location is wrong, and we could stop here. For the purpose of this discussion, we will carry on through the remaining steps.
[math] P_{ux} = -P_{u}\sin\left( \theta \right) = 0\,kip [math]
[math] P_{uy} = -P_{u}\cos\left( \theta \right) = 0\,kip [math]
[math] M_{u} = -P_{ux}\left( Y_{CG} + e_{y} – Y_{IC} \right) + P_{uy} \left( X_{CG} + e_{x} – X_{IC} \right) = 0\,kip\text{-}in [math]
Since,
[math] P_{ux} \neq \sum R_{x} [math]
[math] P_{uy} \neq \sum R_{y} [math]
[math] M_{u} \neq \sum M_{r} [math]
The applied moment is zero while the bolt reactions sum to 203.244 kip-in, so equilibrium is not satisfied. The assumed location of the I.C. is incorrect, and we move to the next assumed location.
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Second Iteration
For our second iteration, let us assume the I.C. is located at the coordinates shown below. Because the load is a downward shear applied to the right of the group, the true IC lies to the left of the centroid, so the trial moves in that direction.
Assume
[math] X_{IC} = 0.062\,in [math]
[math] Y_{IC} = 1.5\,in [math]
Then repeat the steps from the first iteration. The table below shows the coordinates, the distance of each bolt from the assumed I.C., and the corresponding deformation with respect to that distance.
| Bolt ID | X (in) | Y (in) | cx (in) | cy (in) | c (in) | [mathin] \Delta [mathin] (in) |
| 1 | 0 | 0 | -0.062 | -1.5 | 1.501 | 0.155 |
| 2 | 0 | 3 | -0.062 | 1.5 | 1.501 | 0.155 |
| 3 | 3 | 0 | 2.938 | -1.5 | 3.299 | 0.34 |
| 4 | 3 | 3 | 2.938 | 1.5 | 3.299 | 0.34 |
This time the bolts are not all the same distance from the IC, so only the far pair reaches the 0.34 in limit. The near pair deforms less and therefore carries less load.
Note that the calculated centroid of the bolt group is unchanged, since the bolt coordinates have not moved.
[math] X_{CG} = 1.5\,in [math]
[math] Y_{CG} = 1.5\,in [math]
Then we calculate the reactions along x, the reactions along y, and the corresponding moment. The values are tabulated below.
| Bolt ID | R (kip) | Rx (kip) | Ry (kip) | Mr (kip-in) |
| 1 | 21.4 | 21.4 | -0.9 | 32.1 |
| 2 | 21.4 | -21.4 | -0.9 | 32.1 |
| 3 | 23.9 | 10.9 | 21.3 | 79.0 |
| 4 | 23.9 | -10.9 | 21.3 | 79.0 |
| ⅀Rx = 0 | ⅀Ry = 41 | ⅀Mr = 222 |
Next, we determine the resultant load of all the reactions along x and y.
[math] P_{u} = \sqrt{ {\left( \sum R_{x} \right)}^{2} + {\left( \sum R_{y} \right)}^{2} } [math]
[math] P_{u} = \sqrt{ {\left( 0\,kip \right)}^{2} + {\left( 40.703\,kip \right)}^{2} } [math]
[math] P_{u} = 40.703\,kip [math]
Then the components of the resultant load, based on the given [mathin] \theta [mathin], are shown below.
[math] P_{ux} = -P_{u}\sin\left( \theta \right) = -41\,kip \times \sin\left( 0\,deg \right) = 0\,kip [math]
[math] P_{uy} = -P_{u}\cos\left( \theta \right) = -41\,kip \times \cos\left( 0\,deg \right) = -41\,kip [math]
We then use these components to solve for the moment load about the assumed I.C.
[math] M_{u} = -P_{ux} \left( Y_{CG} + e_{y} – Y_{IC} \right) + P_{uy} \left( X_{CG} + e_{x} – X_{IC} \right) [math]
[math] M_{u} = -0\,kip \left( 1.5\,in + 0\,in – 1.5\,in \right) + \left( -41\,kip \right) \left( 1.5\,in + 4\,in – 0.062\,in \right) [math]
[math] M_{u} = -222\,kip\text{-}in [math]
Next, let us compare the calculated [mathin] P_{ux} [mathin], [mathin] P_{uy} [mathin] and [mathin] M_{u} [mathin] against the reactions of the bolt group.
[math] P_{ux} \approx -\sum R_{x} [math]
[math] P_{uy} \approx -\sum R_{y} [math]
[math] M_{u} \approx -\sum M_{r} [math]
Numerically that is 0 against 0, -41 kip against -41 kip, and -222 kip-in against -222 kip-in. The left hand side is close enough to the right hand side that we can take the assumed location of the I.C. as correct.
In practice the solver keeps moving the trial IC until these three residuals fall inside a tolerance, rather than stopping on a visual match.
Solving for C coefficient
Once the I.C. location is determined, we can get the bolt group coefficient C with the formula below.
[math] C = \frac{P_{u}}{\phi R_{n}} = \frac{40.703\,kip}{24.4\,kip} = 1.668 [math]
The coefficient means the group carries 1.668 times the strength of a single bolt. Note that this is well below the four bolts present, because the eccentricity costs the group most of its nominal capacity. It is also where the choice of [mathin] R_{ult} [mathin] cancels, which is why any consistent value could be used at the start.
Free Bolt Group Calculator
See how we design our bolted connections with this approach in the connection design module. Pick a bolted connection, set the bolt layout and the eccentricity, and the report gives the IC location and the C coefficient along with every other check. It runs in the browser, with nothing to install.
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