In stalen verbindingsuitvoering, bouten 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. De directe belastingen worden verdeeld over het totale aantal bouten, 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, het levert vaak overontworpen boutverbindingen op.
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. Om de inelastische benadering te doen, the instantaneous centre of rotation (ICOR) methode met behulp van iteraties is de beste manier.
In dit artikel, we demonstrate how to calculate the strength of a geboute verbinding volgens de ICOR-methode. The reactions per bolt are calculated using Equation (7-1) op pagina 7-7 van de AISC 15e editie-handleiding. That is then used to check whether the assumed location of the instantaneous centre of the bolt group is correct. Uiteindelijk, zodra we de juiste IC-locatie hebben, 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. De SkyCiv Bolt Group-oplosser 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.
De eigenschappen van de boutgroep verkrijgen
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 rechts van de boutgroep. The angle from the vertical is zero and the eccentricity along the y axis is zero.
[wiskunde] V_{u} = 10\,kips [wiskunde]
[wiskunde] \theta = 0\,deg [wiskunde]
[wiskunde] e_{X} = 4\,in [wiskunde]
[wiskunde] e_{j} = 0\,in [wiskunde]
The first thing to do is to get the coordinates of all the bolts in the group. Visual guides and tables are highly recommended.
| Winkel-ID | X (in) | EN (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 hebben de onderstaande formule nodig.
Laten [mathin] n [mathin] = totaal aantal bouten
[wiskunde] X_{CG} = frac{\som X}{n} [wiskunde]
[wiskunde] J_{CG} = frac{\som Y}{n} [wiskunde]
Vervolgens, onze oplossing is::
[wiskunde] X_{CG} = frac{\som X}{n} = frac{0\,in + 0\,in + 3\,in + 3\,in}{4} = 1.5\,in [wiskunde]
[wiskunde] J_{CG} = frac{\som Y}{n} = frac{0\,in + 3\,in + 0\,in + 3\,in}{4} = 1.5\,in [wiskunde]
Neem de locatie van de I.C.
Na het krijgen van het zwaartepunt, we assume the location of the instantaneous centre (IC). Als eerste poging, we can assume the IC sits at the geometric centroid of the bolt group.
Dus, aannemen
[wiskunde] X_{IC} = X_{CG} = 1.5\,in [wiskunde]
[wiskunde] J_{IC} = J_{CG} = 1.5\,in [wiskunde]
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.
| Winkel-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 |
Waarbij,
[wiskunde] c_{X} = X_{ik} – X_{IC} [wiskunde]
[wiskunde] c_{j} = J_{ik} – J_{IC} [wiskunde]
[wiskunde] c = sqrt{ {\links( c_{X} \Rechtsaf)}^{2} + {\links( c_{j} \Rechtsaf)}^{2} } [wiskunde]
Voor bout nr. 1, onze oplossing is:
[wiskunde] c_{X} = 0\,in – 1.5\,in = -1.5\,in [wiskunde]
[wiskunde] c_{j} = 0\,in – 1.5\,in = -1.5\,in [wiskunde]
[wiskunde] c = sqrt{ {\links( -1.5\,in rechts)}^{2} + {\links( -1.5\,in rechts)}^{2} } = 2.121\,in [wiskunde]
Bereken de vervorming per bout tov afstand van IC
After getting the bolt distances from the assumed IC location, we calculate the deformation of each bolt as a function of its distance.
De maximale vervorming per bout, [mathin] \Delta_{max} = 0.34\,in [mathin], is gebaseerd op experimentele gegevens voor een ASTM-bout zoals beschreven op de AISC-pagina 7-8. By linear proportion, en instelling [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]. De vergelijking voor het krijgen van [mathin] \Delta [mathin] wordt hieronder weergegeven:.
[wiskunde] \Delta_{1} = 0.34\,in \times \left( \frac{c}{c_{max}} \Rechtsaf) [wiskunde]
Voor bout nr. 1, de vervorming is
[wiskunde] \Delta_{1} = 0.34\,in \times \left( \frac{2.121\,in}{2.121\,in} \Rechtsaf) = 0.34\,in [wiskunde]
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.
| Winkel-ID | [mathin] \Delta [mathin] (in) |
| 1 | 0.34 |
| 2 | 0.34 |
| 3 | 0.34 |
| 4 | 0.34 |
Krijg de reacties per bout
Zodra we de vervorming per bout hebben:, we use AISC 15th Ed. Eq (7-1) om de reacties per bout te krijgen.
[wiskunde] R = R_{ult} \links( 1 – e^{-10\Delta} \Rechtsaf)^{0.55} [wiskunde]
De [mathin] R_{ult} [mathin] in de vergelijking is de veronderstelde uiteindelijke belasting op een bout, die we kunnen instellen als de afschuifsterkte van de bout.
[wiskunde] R_{ult} = phi R_{n} [wiskunde]
Voor ons voorbeeld, 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.
Voor bout nr. 1, de berekende reactie is
[wiskunde] R = R_{ult} \links( 1 – e^{-10\Delta} \Rechtsaf)^{0.55} [wiskunde]
[wiskunde] R = 24.4\,kip \left( 1 – e^{-10 \keer links( 0.34\,in rechts)} \Rechtsaf)^{0.55} [wiskunde]
[wiskunde] R = 23.949\,kip [wiskunde]
Voor de rest van de bouten, de berekende reacties zijn als volgt:. The components of the bolt reaction [mathin] R [mathin] langs x en y worden ook getoond.
| Winkel-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. Voor Bout No.1, the solutions for the x and y components are shown below.
[wiskunde] R_{X} = -R links( \frac{c_{j}}{c} \Rechtsaf) = -23.949 \keer links( \frac{-1.5\,in}{2.121\,in} \Rechtsaf) = 16.937\,kip [wiskunde]
[wiskunde] R_{j} = R links( \frac{c_{X}}{c} \Rechtsaf) = 23.949 \keer links( \frac{-1.5\,in}{2.121\,in} \Rechtsaf) = -16.937\,kip [wiskunde]
Next we need the moment each bolt reaction produces about the IC. Use the components [mathin] R_{X} [mathin] en [mathin] R_{j} [mathin] with the lever arms [mathin] c_{j} [mathin] en [mathin] c_{X} [mathin].
[wiskunde] M_{r} = -R_{X} c_{j} + R_{j} c_{X} [wiskunde]
Voor Bout No.1, het moment dat de reactie over de IC is
[wiskunde] M_{r} = -16.937\,kip \times \left( -1.5\,in rechts) + \links( -16.937\,kip \right) \keer links( -1.5\,in rechts) [wiskunde]
[wiskunde] M_{r} = 50.811\,kip\text{-}in [wiskunde]
As a check, the same moment is [mathin] R \times c = 23.949 \keer 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.
| Winkel-ID | Dhr (kip-in) |
| 1 | 50.811 |
| 2 | 50.811 |
| 3 | 50.811 |
| 4 | 50.811 |
| ⅀Mr = 203.244 |
De IC-locatie verifiëren
Nu we de afschuif- en momentreacties per bout hebben, 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.
Uit het vorige gedeelte, we calculated that
[wiskunde] \som R_{X} = 0\,kip [wiskunde]
en
[wiskunde] \som R_{j} = 0\,kip [wiskunde]
Dus,
[wiskunde] P_{u} = sqrt{ {\links( \som R_{X} \Rechtsaf)}^{2} + {\links( \som R_{j} \Rechtsaf)}^{2} } = 0\,kip [wiskunde]
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.
[wiskunde] P_{ux} = -P_{u}\zondelinks( \theta rechts) = 0\,kip [wiskunde]
[wiskunde] P_{uy} = -P_{u}\coslinks( \theta rechts) = 0\,kip [wiskunde]
[wiskunde] M_{u} = -P_{ux}\links( J_{CG} + e_{j} – J_{IC} \Rechtsaf) + P_{uy} \links( X_{CG} + e_{X} – X_{IC} \Rechtsaf) = 0\,kip\text{-}in [wiskunde]
Sinds,
[wiskunde] P_{ux} \neq som R_{X} [wiskunde]
[wiskunde] P_{uy} \neq som R_{j} [wiskunde]
[wiskunde] M_{u} \ik ben niet M_{r} [wiskunde]
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 onjuist, and we move to the next assumed location.
SkyCiv has the bolt group calculation built into its connection design modules. Wilt u onze software voor het ontwerpen van verbindingen uitproberen?
Tweede iteratie
Voor onze tweede iteratie, let us assume the I.C. bevindt zich op de onderstaande coördinaten. 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.
Aannemen
[wiskunde] X_{IC} = 0.062\,in [wiskunde]
[wiskunde] J_{IC} = 1.5\,in [wiskunde]
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.
| Winkel-ID | X (in) | EN (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.
Merk op dat het berekende zwaartepunt van de bout groep is unchanged, since the bolt coordinates have not moved.
[wiskunde] X_{CG} = 1.5\,in [wiskunde]
[wiskunde] J_{CG} = 1.5\,in [wiskunde]
Then we calculate the reactions along x, the reactions along y, en het bijbehorende moment. De waarden zijn hieronder weergegeven:.
| Winkel-ID | R (kip) | Rx (kip) | Ry (kip) | Dhr (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 |
De volgende, we bepalen de resulterende belasting van alle reacties langs x en y.
[wiskunde] P_{u} = sqrt{ {\links( \som R_{X} \Rechtsaf)}^{2} + {\links( \som R_{j} \Rechtsaf)}^{2} } [wiskunde]
[wiskunde] P_{u} = sqrt{ {\links( 0\,kip \right)}^{2} + {\links( 40.703\,kip \right)}^{2} } [wiskunde]
[wiskunde] P_{u} = 40.703\,kip [wiskunde]
Then the components of the resultant load, based on the given [mathin] \theta [mathin], are shown below.
[wiskunde] P_{ux} = -P_{u}\zondelinks( \theta rechts) = -41\,kip \times \sin\left( 0\,graden rechts) = 0\,kip [wiskunde]
[wiskunde] P_{uy} = -P_{u}\coslinks( \theta rechts) = -41\,kip \times \cos\left( 0\,graden rechts) = -41\,kip [wiskunde]
We then use these components to solve for the moment load about the assumed I.C.
[wiskunde] M_{u} = -P_{ux} \links( J_{CG} + e_{j} – J_{IC} \Rechtsaf) + P_{uy} \links( X_{CG} + e_{X} – X_{IC} \Rechtsaf) [wiskunde]
[wiskunde] M_{u} = -0\,kip links( 1.5\,in + 0\,in – 1.5\,in rechts) + \links( -41\,kip \right) \links( 1.5\,in + 4\,in – 0.062\,in rechts) [wiskunde]
[wiskunde] M_{u} = -222\,kip\text{-}in [wiskunde]
De volgende, laten we de berekende vergelijken [mathin] P_{ux} [mathin], [mathin] P_{uy} [mathin] en [mathin] M_{u} [mathin] against the reactions of the bolt group.
[wiskunde] P_{ux} \approx -\sum R_{X} [wiskunde]
[wiskunde] P_{uy} \approx -\sum R_{j} [wiskunde]
[wiskunde] M_{u} \approx -\sum M_{r} [wiskunde]
Numerically that is 0 against 0, -41 kip against -41 kip, en -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.
Oplossen voor C-coëfficiënt
Zodra de I.C. locatie is bepaald, we can get the bolt group coefficient C with the formula below.
[wiskunde] C = frac{P_{u}}{\phi R_{n}} = frac{40.703\,kip}{24.4\,kip} = 1.668 [wiskunde]
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.
Gratis boutgroepcalculator
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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