En diseño de conexión de acero, tornillos 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. Las cargas directas se distribuyen entre el número total de pernos, 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, a menudo produce conexiones atornilladas sobrediseñadas.
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. Para hacer el enfoque inelástico, the instantaneous centre of rotation (ICOR) método usando iteraciones es la mejor manera.
En este artículo, we demonstrate how to calculate the strength of a conexión atornillada utilizando el método ICOR. The reactions per bolt are calculated using Equation (7-1) en la página 7-7 del Manual de la 15ª Edición del AISC. That is then used to check whether the assumed location of the instantaneous centre of the bolt group is correct. Finalmente, una vez que tengamos la ubicación correcta del 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. El Solucionador de grupo 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.
Obtener las propiedades del grupo de tornillos
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 pulgadas a la derecha del grupo de tornillos. The angle from the vertical is zero and the eccentricity along the y axis is zero.
[math] V_{tu} = 10\,kips [math]
[math] \theta = 0\,deg [math]
[math] mi_{x} = 4\,in [math]
[math] mi_{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.
| Identificación de la tienda | 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, necesitamos la siguiente fórmula.
Dejar [mathin] norte [mathin] = número total de pernos
[math] X_{C.G.} = frac{\suma X}{norte} [math]
[math] Y_{C.G.} = frac{\suma Y}{norte} [math]
Luego, nuestra solución es:
[math] X_{C.G.} = frac{\suma X}{norte} = frac{0\,in + 0\,in + 3\,in + 3\,in}{4} = 1.5\,in [math]
[math] Y_{C.G.} = frac{\suma Y}{norte} = frac{0\,in + 3\,in + 0\,in + 3\,in}{4} = 1.5\,in [math]
Asumir la ubicación del I.C..
Después de obtener el centroide, we assume the location of the instantaneous centre (CI). como primer intento, we can assume the IC sits at the geometric centroid of the bolt group.
Entonces, asumir
[math] X_{CI} = X_{C.G.} = 1.5\,in [math]
[math] Y_{CI} = Y_{C.G.} = 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.
| Identificación de la tienda | 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 |
Dónde,
[math] C_{x} = X_{i} – X_{CI} [math]
[math] C_{y} = Y_{i} – Y_{CI} [math]
[math] c = sqrt{ {\izquierda( C_{x} \verdad)}^{2} + {\izquierda( C_{y} \verdad)}^{2} } [math]
Para perno No.. 1, nuestra solución es
[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{ {\izquierda( -1.5\,en la derecha)}^{2} + {\izquierda( -1.5\,en la derecha)}^{2} } = 2.121\,in [math]
Calcule la deformación por perno respecto a la distancia desde 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 deformación máxima por perno., [mathin] \Delta_{max} = 0.34\,in [mathin], se basa en datos experimentales para un perno ASTM como se describe en la página AISC 7-8. By linear proportion, y configuración [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]. La ecuación para obtener [mathin] \Delta [mathin] se muestra a continuación.
[math] \Delta_{1} = 0.34\,in \times \left( \frac{c}{C_{max}} \verdad) [math]
Para perno No.. 1, la deformación es
[math] \Delta_{1} = 0.34\,in \times \left( \frac{2.121\,in}{2.121\,in} \verdad) = 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.
| Identificación de la tienda | [mathin] \Delta [mathin] (in) |
| 1 | 0.34 |
| 2 | 0.34 |
| 3 | 0.34 |
| 4 | 0.34 |
Obtener las reacciones por perno
Una vez que tenemos la deformación por tornillo, we use AISC 15th Ed. Eq (7-1) para obtener las reacciones por tornillo.
[math] R = R_{ult} \izquierda( 1 – e ^{-10\Delta} \verdad)^{0.55} [math]
El [mathin] R_{ult} [mathin] en la ecuación es la carga última asumida sobre un perno, que podemos establecer como la resistencia a cortante del perno.
[math] R_{ult} = fi R_{norte} [math]
Para nuestro ejemplo, 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] mas tarde.
Para perno No.. 1, la reacción calculada es
[math] R = R_{ult} \izquierda( 1 – e ^{-10\Delta} \verdad)^{0.55} [math]
[math] R = 24.4\,kip \left( 1 – e ^{-10 \veces left( 0.34\,en la derecha)} \verdad)^{0.55} [math]
[math] R = 23.949\,kip [math]
Para el resto de los tornillos, las reacciones calculadas son las siguientes. The components of the bolt reaction [mathin] R [mathin] a lo largo de x e y también se muestran.
| Identificación de la tienda | 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. Para Perno No.1, the solutions for the x and y components are shown below.
[math] R_{x} = -R izquierda( \frac{C_{y}}{c} \verdad) = -23.949 \veces left( \frac{-1.5\,in}{2.121\,in} \verdad) = 16.937\,kip [math]
[math] R_{y} = R izquierda( \frac{C_{x}}{c} \verdad) = 23.949 \veces left( \frac{-1.5\,in}{2.121\,in} \verdad) = -16.937\,kip [math]
Next we need the moment each bolt reaction produces about the IC. Use the components [mathin] R_{x} [mathin] y [mathin] R_{y} [mathin] with the lever arms [mathin] C_{y} [mathin] y [mathin] C_{x} [mathin].
[math] METRO_{r} = -R_{x} C_{y} + R_{y} C_{x} [math]
Para Perno No.1, el momento en que la reacción sobre el IC es
[math] METRO_{r} = -16.937\,kip \times \left( -1.5\,en la derecha) + \izquierda( -16.937\,kip \right) \veces left( -1.5\,en la derecha) [math]
[math] METRO_{r} = 50.811\,kip\text{-}in [math]
As a check, the same moment is [mathin] R \times c = 23.949 \veces 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.
| Identificación de la tienda | Señor (pollo en) |
| 1 | 50.811 |
| 2 | 50.811 |
| 3 | 50.811 |
| 4 | 50.811 |
| ⅀señor = 203.244 |
Verificación de la ubicación del IC
Ahora que tenemos las reacciones de cortante y momento por tornillo, 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 sección anterior, we calculated that
[math] \suma R_{x} = 0\,kip [math]
y
[math] \suma R_{y} = 0\,kip [math]
Entonces,
[math] el calculo de la resultante es como sigue{tu} = sqrt{ {\izquierda( \suma R_{x} \verdad)}^{2} + {\izquierda( \suma R_{y} \verdad)}^{2} } = 0\,kip [math]
The resulting load is [mathin] el calculo de la resultante es como sigue{tu} = 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] el calculo de la resultante es como sigue{ux} = -P_{tu}\pecado izquierda( \theta derecho) = 0\,kip [math]
[math] el calculo de la resultante es como sigue{uy} = -P_{tu}\porqueizquierda( \theta derecho) = 0\,kip [math]
[math] METRO_{tu} = -P_{ux}\izquierda( Y_{C.G.} + mi_{y} – Y_{CI} \verdad) + el calculo de la resultante es como sigue{uy} \izquierda( X_{C.G.} + mi_{x} – X_{CI} \verdad) = 0\,kip\text{-}in [math]
Ya que,
[math] el calculo de la resultante es como sigue{ux} \neq sum R_{x} [math]
[math] el calculo de la resultante es como sigue{uy} \neq sum R_{y} [math]
[math] METRO_{tu} \no soy m_{r} [math]
The applied moment is zero while the bolt reactions sum to 203.244 pollo en, so equilibrium is not satisfied. The assumed location of the I.C. Es incorrecto, and we move to the next assumed location.
SkyCiv has the bolt group calculation built into its connection design modules. Quiere probar nuestro software de diseño de conexiones?
Segunda iteración
Para nuestra segunda iteración, let us assume the I.C. se encuentra en las coordenadas que se muestran a continuación. 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.
Asumir
[math] X_{CI} = 0.062\,in [math]
[math] Y_{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.
| Identificación de la tienda | 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.
Tenga en cuenta que el centroide calculado de la grupo de pernos is unchanged, since the bolt coordinates have not moved.
[math] X_{C.G.} = 1.5\,in [math]
[math] Y_{C.G.} = 1.5\,in [math]
Then we calculate the reactions along x, the reactions along y, y el momento correspondiente. Los valores se tabulan a continuación..
| Identificación de la tienda | R (kip) | Rx (kip) | Ry (kip) | Señor (pollo en) |
| 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 | ⅀señor = 222 |
próximo, determinamos la carga resultante de todas las reacciones a lo largo de x e y.
[math] el calculo de la resultante es como sigue{tu} = sqrt{ {\izquierda( \suma R_{x} \verdad)}^{2} + {\izquierda( \suma R_{y} \verdad)}^{2} } [math]
[math] el calculo de la resultante es como sigue{tu} = sqrt{ {\izquierda( 0\,kip \right)}^{2} + {\izquierda( 40.703\,kip \right)}^{2} } [math]
[math] el calculo de la resultante es como sigue{tu} = 40.703\,kip [math]
Then the components of the resultant load, based on the given [mathin] \theta [mathin], are shown below.
[math] el calculo de la resultante es como sigue{ux} = -P_{tu}\pecado izquierda( \theta derecho) = -41\,kip \times \sin\left( 0\,grado derecho) = 0\,kip [math]
[math] el calculo de la resultante es como sigue{uy} = -P_{tu}\porqueizquierda( \theta derecho) = -41\,kip \times \cos\left( 0\,grado derecho) = -41\,kip [math]
We then use these components to solve for the moment load about the assumed I.C.
[math] METRO_{tu} = -P_{ux} \izquierda( Y_{C.G.} + mi_{y} – Y_{CI} \verdad) + el calculo de la resultante es como sigue{uy} \izquierda( X_{C.G.} + mi_{x} – X_{CI} \verdad) [math]
[math] METRO_{tu} = -0\,kip izquierda( 1.5\,in + 0\,in – 1.5\,en la derecha) + \izquierda( -41\,kip \right) \izquierda( 1.5\,in + 4\,in – 0.062\,en la derecha) [math]
[math] METRO_{tu} = -222\,kip\text{-}in [math]
próximo, comparemos el calculado [mathin] el calculo de la resultante es como sigue{ux} [mathin], [mathin] el calculo de la resultante es como sigue{uy} [mathin] y [mathin] METRO_{tu} [mathin] against the reactions of the bolt group.
[math] el calculo de la resultante es como sigue{ux} \approx -\sum R_{x} [math]
[math] el calculo de la resultante es como sigue{uy} \approx -\sum R_{y} [math]
[math] METRO_{tu} \approx -\sum M_{r} [math]
Numerically that is 0 against 0, -41 kip against -41 kip, y -222 kip-in against -222 pollo en. 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.
Resolviendo para el coeficiente C
Una vez que el I.C.. la ubicación está determinada, we can get the bolt group coefficient C with the formula below.
[math] C = frac{el calculo de la resultante es como sigue{tu}}{\fi R_{norte}} = 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.
Calculadora de grupo de tornillos libres
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. Se ejecuta en el navegador., sin nada que instalar.
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