Dans la conception de connexion en acier, boulons 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. Les charges directes sont réparties sur le nombre total de boulons, 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, il donne souvent des assemblages boulonnés surdimensionnés.
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. Pour faire l'approche inélastique, the instantaneous centre of rotation (ICOR) méthode utilisant des itérations est la meilleure façon.
Dans cet article, we demonstrate how to calculate the strength of a connexion boulonnée en utilisant la méthode ICOR. The reactions per bolt are calculated using Equation (7-1) sur la page 7-7 du manuel AISC 15e édition. That is then used to check whether the assumed location of the instantaneous centre of the bolt group is correct. Ensuite, une fois que nous avons le bon emplacement IC, 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 (CI) is trial and error. With a computer solver the IC of a bolt group is found by programmed iterations instead. Ce logiciel Solveur de groupe SkyCiv Bolt 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.
Obtenir les propriétés du groupe de boulons
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 pouces à droite du groupe de boulons. 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.
| Identifiant du magasin | 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, nous avons besoin de la formule ci-dessous.
Laisser [mathin] n [mathin] = nombre total de boulons
[math] X_{CG} = frac{\somme X}{n} [math]
[math] O_{CG} = frac{\somme Y}{n} [math]
ensuite, notre solution est:
[math] X_{CG} = frac{\somme X}{n} = frac{0\,in + 0\,in + 3\,in + 3\,in}{4} = 1.5\,in [math]
[math] O_{CG} = frac{\somme Y}{n} = frac{0\,in + 3\,in + 0\,in + 3\,in}{4} = 1.5\,in [math]
Supposons l'emplacement de l'I.C.
Après avoir obtenu le centroïde, we assume the location of the instantaneous centre (CI). Comme un premier essai, we can assume the IC sits at the geometric centroid of the bolt group.
Donc, présumer
[math] X_{CI} =X_{CG} = 1.5\,in [math]
[math] O_{CI} = 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.
| Identifiant du magasin | 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 |
Où,
[math] c_{X} =X_{je} – X_{CI} [math]
[math] c_{Y} = Y_{je} – O_{CI} [math]
[math] c = sqrt{ {\la gauche( c_{X} \droite)}^{2} + {\la gauche( c_{Y} \droite)}^{2} } [math]
Pour Boulon Non. 1, notre solution est
[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{ {\la gauche( -1.5\,À droite)}^{2} + {\la gauche( -1.5\,À droite)}^{2} } = 2.121\,in [math]
Calculer la déformation par boulon par rapport à la distance de IC
After getting the bolt distances from the assumed IC location, we calculate the deformation of each bolt as a function of its distance.
La déformation maximale par boulon, [mathin] \Delta_{max} = 0.34\,in [mathin], est basé sur des données expérimentales pour un boulon ASTM tel que décrit dans la page AISC 7-8. By linear proportion, et réglage [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]. L'équation pour obtenir [mathin] \Delta [mathin] est montré ci-dessous.
[math] \Delta_{1} = 0.34\,in \times \left( \frac{c}{c_{max}} \droite) [math]
Pour Boulon Non. 1, la déformation est
[math] \Delta_{1} = 0.34\,in \times \left( \frac{2.121\,in}{2.121\,in} \droite) = 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.
| Identifiant du magasin | [mathin] \Delta [mathin] (in) |
| 1 | 0.34 |
| 2 | 0.34 |
| 3 | 0.34 |
| 4 | 0.34 |
Obtenez les réactions par boulon
Une fois que nous avons la déformation par boulon, we use AISC 15th Ed. Eq (7-1) pour obtenir les réactions par boulon.
[math] R = R_{ultime} \la gauche( 1 – e^{-10\Delta} \droite)^{0.55} [math]
Ce logiciel [mathin] R_{ultime} [mathin] dans l'équation est la charge ultime supposée sur un boulon, que nous pouvons définir comme la résistance au cisaillement des boulons.
[math] R_{ultime} = phi R_{n} [math]
Dans notre exemple, 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] plus tard.
Pour Boulon Non. 1, la réaction calculée est
[math] R = R_{ultime} \la gauche( 1 – e^{-10\Delta} \droite)^{0.55} [math]
[math] R = 24.4\,kip \left( 1 – e^{-10 \fois gauche( 0.34\,À droite)} \droite)^{0.55} [math]
[math] R = 23.949\,kip [math]
Pour le reste des boulons, les réactions calculées sont les suivantes. The components of the bolt reaction [mathin] R [mathin] le long de x et y sont également affichés.
| Identifiant du magasin | 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. Pour Boulon No.1, the solutions for the x and y components are shown below.
[math] R_{X} = -R gauche( \frac{c_{Y}}{c} \droite) = -23.949 \fois gauche( \frac{-1.5\,in}{2.121\,in} \droite) = 16.937\,kip [math]
[math] R_{Y} = Rgauche( \frac{c_{X}}{c} \droite) = 23.949 \fois gauche( \frac{-1.5\,in}{2.121\,in} \droite) = -16.937\,kip [math]
Next we need the moment each bolt reaction produces about the IC. Use the components [mathin] R_{X} [mathin] et [mathin] R_{Y} [mathin] with the lever arms [mathin] c_{Y} [mathin] et [mathin] c_{X} [mathin].
[math] M_{r} = -R_{X} c_{Y} + R_{Y} c_{X} [math]
Pour Boulon No.1, le moment de la réaction sur le CI est
[math] M_{r} = -16.937\,kip \times \left( -1.5\,À droite) + \la gauche( -16.937\,kip \right) \fois gauche( -1.5\,À droite) [math]
[math] M_{r} = 50.811\,kip\text{-}in [math]
As a check, the same moment is [mathin] R \times c = 23.949 \fois 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.
| Identifiant du magasin | M (poulet dans) |
| 1 | 50.811 |
| 2 | 50.811 |
| 3 | 50.811 |
| 4 | 50.811 |
| ⅀Mr = 203.244 |
Vérification de l'emplacement du CI
Maintenant que nous avons les réactions de cisaillement et de moment par boulon, 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.
De la section précédente, we calculated that
[math] \somme R_{X} = 0\,kip [math]
et
[math] \somme R_{Y} = 0\,kip [math]
Donc,
[math] P_{u} = sqrt{ {\la gauche( \somme R_{X} \droite)}^{2} + {\la gauche( \somme R_{Y} \droite)}^{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}\péchégauche( \thêta droit) = 0\,kip [math]
[math] P_{ouais} = -P_{u}\cosgauche( \thêta droit) = 0\,kip [math]
[math] M_{u} = -P_{ux}\la gauche( O_{CG} + e_{Y} – O_{CI} \droite) + P_{ouais} \la gauche( X_{CG} + e_{X} – X_{CI} \droite) = 0\,kip\text{-}in [math]
Puisque,
[math] P_{ux} \neq somme R_{X} [math]
[math] P_{ouais} \neq somme R_{Y} [math]
[math] M_{u} \je ne suis pas M_{r} [math]
The applied moment is zero while the bolt reactions sum to 203.244 poulet dans, so equilibrium is not satisfied. The assumed location of the I.C. est incorrect, and we move to the next assumed location.
SkyCiv has the bolt group calculation built into its connection design modules. Vous voulez essayer notre logiciel de conception de connexion?
Deuxième itération
Pour notre deuxième itération, let us assume the I.C. est situé aux coordonnées indiquées ci-dessous. 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.
Présumer
[math] X_{CI} = 0.062\,in [math]
[math] O_{CI} = 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.
| Identifiant du magasin | 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.
Notez que le centre de gravité calculé du groupe de boulons is unchanged, since the bolt coordinates have not moved.
[math] X_{CG} = 1.5\,in [math]
[math] O_{CG} = 1.5\,in [math]
Then we calculate the reactions along x, the reactions along y, et l'instant correspondant. Les valeurs sont tabulées ci-dessous.
| Identifiant du magasin | R (kip) | Rx (kip) | Ry (kip) | M (poulet dans) |
| 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 |
Prochain, nous déterminons la charge résultante de toutes les réactions le long de x et y.
[math] P_{u} = sqrt{ {\la gauche( \somme R_{X} \droite)}^{2} + {\la gauche( \somme R_{Y} \droite)}^{2} } [math]
[math] P_{u} = sqrt{ {\la gauche( 0\,kip \right)}^{2} + {\la gauche( 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] \thêta [mathin], are shown below.
[math] P_{ux} = -P_{u}\péchégauche( \thêta droit) = -41\,kip \times \sin\left( 0\,degré droit) = 0\,kip [math]
[math] P_{ouais} = -P_{u}\cosgauche( \thêta droit) = -41\,kip \times \cos\left( 0\,degré droit) = -41\,kip [math]
We then use these components to solve for the moment load about the assumed I.C.
[math] M_{u} = -P_{ux} \la gauche( O_{CG} + e_{Y} – O_{CI} \droite) + P_{ouais} \la gauche( X_{CG} + e_{X} – X_{CI} \droite) [math]
[math] M_{u} = -0\,kip gauche( 1.5\,in + 0\,in – 1.5\,À droite) + \la gauche( -41\,kip \right) \la gauche( 1.5\,in + 4\,in – 0.062\,À droite) [math]
[math] M_{u} = -222\,kip\text{-}in [math]
Prochain, comparons le calcul [mathin] P_{ux} [mathin], [mathin] P_{ouais} [mathin] et [mathin] M_{u} [mathin] against the reactions of the bolt group.
[math] P_{ux} \approx -\sum R_{X} [math]
[math] P_{ouais} \approx -\sum R_{Y} [math]
[math] M_{u} \approx -\sum M_{r} [math]
Numerically that is 0 against 0, -41 kip against -41 kip, et -222 kip-in against -222 poulet dans. 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.
Résolution du coefficient C
Une fois l'I.C.. l'emplacement est déterminé, 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_{ultime} [mathin] cancels, which is why any consistent value could be used at the start.
Calculateur de groupe de boulons gratuit
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. Il s'exécute dans le navigateur, with nothing to install.
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