In Stahlverbindungsausführung, Schrauben 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. Die direkten Lasten werden auf die Gesamtzahl der Schrauben verteilt, 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, es führt oft zu überdimensionierten Schraubverbindungen.
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. Um den unelastischen Ansatz zu machen, the instantaneous centre of rotation (ICOR) Methode mit Iterationen ist der beste Weg.
In diesem Artikel, we demonstrate how to calculate the strength of a Schraubverbindung mit der ICOR-Methode. The reactions per bolt are calculated using Equation (7-1) Auf Seite 7-7 des AISC-Handbuchs, 15. Ausgabe. That is then used to check whether the assumed location of the instantaneous centre of the bolt group is correct. Schließlich, sobald wir den richtigen IC-Standort haben, 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. Mit der 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.
Abrufen der Schraubengruppeneigenschaften
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 Zoll rechts von der Schraubengruppe. The angle from the vertical is zero and the eccentricity along the y axis is zero.
[Mathematik] V_{u} = 10\,kips [Mathematik]
[Mathematik] \theta = 0\,deg [Mathematik]
[Mathematik] e_{x} = 4\,in [Mathematik]
[Mathematik] e_{j} = 0\,in [Mathematik]
The first thing to do is to get the coordinates of all the bolts in the group. Visual guides and tables are highly recommended.
| Shop-ID | X. (im) | UND (im) |
| 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, Wir brauchen die folgende Formel.
Lassen [mathin] n [mathin] = Gesamtzahl der Schrauben
[Mathematik] X_{CG} = frac{\Summe X}{n} [Mathematik]
[Mathematik] Y_{CG} = frac{\Summe Y}{n} [Mathematik]
Dann, Unsere Lösung ist:
[Mathematik] X_{CG} = frac{\Summe X}{n} = frac{0\,im + 0\,im + 3\,im + 3\,im}{4} = 1.5\,in [Mathematik]
[Mathematik] Y_{CG} = frac{\Summe Y}{n} = frac{0\,im + 3\,im + 0\,im + 3\,im}{4} = 1.5\,in [Mathematik]
Angenommen, der Standort des I.C.
Nachdem Sie den Schwerpunkt erhalten haben, we assume the location of the instantaneous centre (IC). Als erster Versuch, we can assume the IC sits at the geometric centroid of the bolt group.
So, davon ausgehen
[Mathematik] X_{IC} = X_{CG} = 1.5\,in [Mathematik]
[Mathematik] Y_{IC} = Y_{CG} = 1.5\,in [Mathematik]
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.
| Shop-ID | cx (im) | cy (im) | c (im) |
| 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 |
Wo,
[Mathematik] c_{x} = X_{ich} – X_{IC} [Mathematik]
[Mathematik] c_{j} = Y_{ich} – Y_{IC} [Mathematik]
[Mathematik] c = sqrt{ {\links( c_{x} \richtig)}^{2} + {\links( c_{j} \richtig)}^{2} } [Mathematik]
Für Schraube Nr. 1, Unsere Lösung ist
[Mathematik] c_{x} = 0\,in – 1.5\,in = -1.5\,in [Mathematik]
[Mathematik] c_{j} = 0\,in – 1.5\,in = -1.5\,in [Mathematik]
[Mathematik] c = sqrt{ {\links( -1.5\,in Recht)}^{2} + {\links( -1.5\,in Recht)}^{2} } = 2.121\,in [Mathematik]
Berechnen Sie die Verformung pro Schraube bzgl. Abstand von IC
After getting the bolt distances from the assumed IC location, we calculate the deformation of each bolt as a function of its distance.
Die maximale Verformung pro Schraube, [mathin] \Delta_{max} = 0.34\,in [mathin], basiert auf experimentellen Daten für eine ASTM-Schraube, wie auf der AISC-Seite beschrieben 7-8. By linear proportion, und Einstellung [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]. Die Gleichung für das Erhalten [mathin] \p-Delta-Effekte [mathin] wird unten gezeigt.
[Mathematik] \Delta_{1} = 0.34\,in \times \left( \frac{c}{c_{max}} \richtig) [Mathematik]
Für Schraube Nr. 1, die Verformung ist
[Mathematik] \Delta_{1} = 0.34\,in \times \left( \frac{2.121\,im}{2.121\,im} \richtig) = 0.34\,in [Mathematik]
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.
| Shop-ID | [mathin] \p-Delta-Effekte [mathin] (im) |
| 1 | 0.34 |
| 2 | 0.34 |
| 3 | 0.34 |
| 4 | 0.34 |
Holen Sie sich die Reaktionen pro Bolzen
Sobald wir die Verformung pro Schraube haben, we use AISC 15th Ed. Gl (7-1) um die Reaktionen pro Schraube zu erhalten.
[Mathematik] R = R_{ult} \links( 1 – e^{-10\p-Delta-Effekte} \richtig)^{0.55} [Mathematik]
Mit der [mathin] R_{ult} [mathin] in der Gleichung ist die angenommene Bruchlast einer Schraube, die wir als Schraubenscherfestigkeit einstellen können.
[Mathematik] R_{ult} = φR_{n} [Mathematik]
Für unser Beispiel, 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] später.
Für Schraube Nr. 1, die berechnete Reaktion ist
[Mathematik] R = R_{ult} \links( 1 – e^{-10\p-Delta-Effekte} \richtig)^{0.55} [Mathematik]
[Mathematik] R = 24.4\,kip \left( 1 – e^{-10 \mal links( 0.34\,in Recht)} \richtig)^{0.55} [Mathematik]
[Mathematik] R = 23.949\,kip [Mathematik]
Für den Rest der Schrauben, die berechneten Reaktionen sind wie folgt. The components of the bolt reaction [mathin] R. [mathin] entlang x und y sind ebenfalls gezeigt.
| Shop-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. Für Schraube Nr. 1, the solutions for the x and y components are shown below.
[Mathematik] R_{x} = -R links( \frac{c_{j}}{c} \richtig) = -23.949 \mal links( \frac{-1.5\,im}{2.121\,im} \richtig) = 16.937\,kip [Mathematik]
[Mathematik] R_{j} = R links( \frac{c_{x}}{c} \richtig) = 23.949 \mal links( \frac{-1.5\,im}{2.121\,im} \richtig) = -16.937\,kip [Mathematik]
Next we need the moment each bolt reaction produces about the IC. Use the components [mathin] R_{x} [mathin] und [mathin] R_{j} [mathin] with the lever arms [mathin] c_{j} [mathin] und [mathin] c_{x} [mathin].
[Mathematik] M_{r} = -R_{x} c_{j} + R_{j} c_{x} [Mathematik]
Für Schraube Nr. 1, Momentan ist die Reaktion auf den IC
[Mathematik] M_{r} = -16.937\,kip \times \left( -1.5\,in Recht) + \links( -16.937\,kip \right) \mal links( -1.5\,in Recht) [Mathematik]
[Mathematik] M_{r} = 50.811\,kip\text{-}im [Mathematik]
As a check, the same moment is [mathin] R \times c = 23.949 \mal 2.121 = 50.8\,kip\text{-}im [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.
| Shop-ID | Herr (Hühnchen) |
| 1 | 50.811 |
| 2 | 50.811 |
| 3 | 50.811 |
| 4 | 50.811 |
| ⅀Herr = 203.244 |
Überprüfung des IC-Standorts
Jetzt haben wir die Scher- und Momentreaktionen pro Schraube, 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.
Aus dem vorherigen Abschnitt, we calculated that
[Mathematik] \Summe R_{x} = 0\,kip [Mathematik]
und
[Mathematik] \Summe R_{j} = 0\,kip [Mathematik]
So,
[Mathematik] P_{u} = Quadrat{ {\links( \Summe R_{x} \richtig)}^{2} + {\links( \Summe R_{j} \richtig)}^{2} } = 0\,kip [Mathematik]
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.
[Mathematik] P_{ux} = -P_{u}\Sündelinks( \Theta richtig) = 0\,kip [Mathematik]
[Mathematik] P_{ui} = -P_{u}\coslinks( \Theta richtig) = 0\,kip [Mathematik]
[Mathematik] M_{u} = -P_{ux}\links( Y_{CG} + e_{j} – Y_{IC} \richtig) + P_{ui} \links( X_{CG} + e_{x} – X_{IC} \richtig) = 0\,kip\text{-}im [Mathematik]
Schon seit,
[Mathematik] P_{ux} \neq sum R_{x} [Mathematik]
[Mathematik] P_{ui} \neq sum R_{j} [Mathematik]
[Mathematik] M_{u} \Ich bin nicht M_{r} [Mathematik]
The applied moment is zero while the bolt reactions sum to 203.244 Hühnchen, so equilibrium is not satisfied. The assumed location of the I.C. ist falsch, and we move to the next assumed location.
SkyCiv has the bolt group calculation built into its connection design modules. Möchten Sie unsere Verbindungsdesign-Software ausprobieren?
Zweite Iteration
Für unsere zweite Iteration, let us assume the I.C. befindet sich an den unten angegebenen Koordinaten. 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.
Davon ausgehen
[Mathematik] X_{IC} = 0.062\,in [Mathematik]
[Mathematik] Y_{IC} = 1.5\,in [Mathematik]
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.
| Shop-ID | X. (im) | UND (im) | cx (im) | cy (im) | c (im) | [mathin] \p-Delta-Effekte [mathin] (im) |
| 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.
Beachten Sie, dass der berechnete Schwerpunkt des Schraubengruppe is unchanged, since the bolt coordinates have not moved.
[Mathematik] X_{CG} = 1.5\,in [Mathematik]
[Mathematik] Y_{CG} = 1.5\,in [Mathematik]
Then we calculate the reactions along x, the reactions along y, und dem entsprechenden Moment. Die Werte sind unten tabelliert.
| Shop-ID | R. (kip) | Rx (kip) | Ry (kip) | Herr (Hühnchen) |
| 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 | ⅀Herr = 222 |
Als nächstes, wir bestimmen die resultierende Belastung aller Reaktionen entlang x und y.
[Mathematik] P_{u} = Quadrat{ {\links( \Summe R_{x} \richtig)}^{2} + {\links( \Summe R_{j} \richtig)}^{2} } [Mathematik]
[Mathematik] P_{u} = Quadrat{ {\links( 0\,kip \right)}^{2} + {\links( 40.703\,kip \right)}^{2} } [Mathematik]
[Mathematik] P_{u} = 40.703\,kip [Mathematik]
Then the components of the resultant load, based on the given [mathin] \theta [mathin], are shown below.
[Mathematik] P_{ux} = -P_{u}\Sündelinks( \Theta richtig) = -41\,kip \times \sin\left( 0\,Grad right) = 0\,kip [Mathematik]
[Mathematik] P_{ui} = -P_{u}\coslinks( \Theta richtig) = -41\,kip \times \cos\left( 0\,Grad right) = -41\,kip [Mathematik]
We then use these components to solve for the moment load about the assumed I.C.
[Mathematik] M_{u} = -P_{ux} \links( Y_{CG} + e_{j} – Y_{IC} \richtig) + P_{ui} \links( X_{CG} + e_{x} – X_{IC} \richtig) [Mathematik]
[Mathematik] M_{u} = -0\,kip links( 1.5\,im + 0\,im – 1.5\,in Recht) + \links( -41\,kip \right) \links( 1.5\,im + 4\,im – 0.062\,in Recht) [Mathematik]
[Mathematik] M_{u} = -222\,kip\text{-}im [Mathematik]
Als nächstes, vergleichen wir die errechneten [mathin] P_{ux} [mathin], [mathin] P_{ui} [mathin] und [mathin] M_{u} [mathin] against the reactions of the bolt group.
[Mathematik] P_{ux} \approx -\sum R_{x} [Mathematik]
[Mathematik] P_{ui} \approx -\sum R_{j} [Mathematik]
[Mathematik] M_{u} \approx -\sum M_{r} [Mathematik]
Numerically that is 0 against 0, -41 kip against -41 kip, und -222 kip-in against -222 Hühnchen. 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.
Auflösen nach dem C-Koeffizienten
Sobald der I.C. Standort bestimmt, we can get the bolt group coefficient C with the formula below.
[Mathematik] C = frac{P_{u}}{\Phi R_{n}} = frac{40.703\,kip}{24.4\,kip} = 1.668 [Mathematik]
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.
Kostenloser Schraubengruppenrechner
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. Es läuft im Browser, with nothing to install.
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